Calorimetry

# Calorimetry ## Calorimetry: Definition of heat capacity and specific heat capacity. Measurement of specific heat capacity, c for a solid using Q = mc∆θ ### 1. Introduction Welcome to our lesson on Calorimetry! Calorimetry is a branch of physics that deals with the measurement of heat. In everyday life, we encounter heat transfer constantly, from boiling water to keeping food warm in a thermos. Understanding how different materials absorb or release heat is crucial in many applications, from engineering to cooking. Today, we will define key terms like heat capacity and specific heat capacity, and learn how to measure the specific heat capacity of a solid. ### 2. Key Concepts #### 2.1 What is Heat? Heat is a form of energy that is transferred between objects or systems due to a temperature difference. It always flows from a region of higher temperature to a region of lower temperature. When an object gains heat, its internal energy increases, often leading to a rise in temperature. #### 2.2 Heat Capacity (C) Heat capacity is a measure of how much heat energy is required to raise the temperature of a *specific object or body* by one degree Celsius (or one Kelvin). * **Definition:** The amount of heat energy required to change the temperature of an object by \(1^{\circ}\text{C}\) or \(1\text{ K}\). * **Formula:** \(C = \frac{Q}{\Delta\theta}\) where: * \(C\) is the heat capacity. * \(Q\) is the heat energy supplied or absorbed. * \(\Delta\theta\) is the change in temperature (\(\theta_{final} - \theta_{initial}\)). * **Units:** Joules per degree Celsius (\(J/^{\circ}\text{C}\)) or Joules per Kelvin (\(J/K\)). * **Key Point:** Heat capacity depends on the mass and material of the object. #### 2.3 Specific Heat Capacity (c) Specific heat capacity is a more fundamental property of a *substance* itself, rather than a particular object. It tells us how much heat energy is needed to raise the temperature of a *unit mass* of a substance by one degree Celsius. * **Definition:** The amount of heat energy required to change the temperature of \(1\text{ kg}\) of a substance by \(1^{\circ}\text{C}\) or \(1\text{ K}\). * **Formula:** \(c = \frac{Q}{m\Delta\theta}\) where: * \(c\) is the specific heat capacity. * \(Q\) is the heat energy supplied or absorbed. * \(m\) is the mass of the substance. * \(\Delta\theta\) is the change in temperature. * **Units:** Joules per kilogram per degree Celsius (\(J/kg^{\circ}\text{C}\)) or Joules per kilogram per Kelvin (\(J/kg\text{ K}\)). * **Key Point:** Specific heat capacity is an intensive property, meaning it does not depend on the amount of substance. It's characteristic of the material itself (e.g., water, copper, aluminum). #### 2.4 Relationship between Heat Capacity (C) and Specific Heat Capacity (c) The heat capacity of an object is simply its mass multiplied by the specific heat capacity of the material it's made from. \[ C = mc \] This means if you know the specific heat capacity of a material and the mass of an object made from that material, you can find its total heat capacity. #### 2.5 Calculating Heat Energy (Q = mc∆θ) The most important formula in calorimetry relates the amount of heat energy transferred (\(Q\)) to the mass (\(m\)), specific heat capacity (\(c\)), and the change in temperature (\(\Delta\theta\)) of a substance: \[ Q = mc\Delta\theta \] This formula allows us to calculate how much heat is gained or lost by a substance when its temperature changes. #### 2.6 Measurement of Specific Heat Capacity (c) for a Solid We can determine the specific heat capacity of a solid material using the principle of conservation of energy, often through a method involving a calorimeter (an insulated container). **Method for a Solid (e.g., using a calorimeter and a heater):** 1. **Mass of Solid:** Measure the mass (\(m_s\)) of the solid material whose specific heat capacity (\(c_s\)) you want to find. 2. **Heat the Solid:** Heat the solid to a known high temperature (\(\theta_h\)), typically by placing it in boiling water for a sufficient time so it reaches \(100^{\circ}\text{C}\). Ensure it is dry. 3. **Calorimeter Setup:** Place a known mass of water (\(m_w\)) into an insulated calorimeter. Measure the mass of the calorimeter (\(m_c\)) and its specific heat capacity (\(c_c\)). Measure the initial temperature of the water and calorimeter (\(\theta_i\)). Ensure \(\theta_i\) is significantly lower than \(\theta_h\). 4. **Transfer Solid:** Quickly transfer the hot solid into the water in the calorimeter. Stir the water gently. 5. **Final Temperature:** Observe the temperature rise until it reaches a maximum steady value (\(\theta_f\)). This is the final temperature of the mixture (solid, water, and calorimeter). **Calculations (Principle of Conservation of Energy):** Assuming no heat loss to the surroundings, the heat lost by the hot solid is equal to the heat gained by the water and the calorimeter. * **Heat lost by the solid:** \(Q_{lost} = m_s c_s (\theta_h - \theta_f)\) * **Heat gained by the water:** \(Q_{water} = m_w c_w (\theta_f - \theta_i)\) * **Heat gained by the calorimeter:** \(Q_{calorimeter} = m_c c_c (\theta_f - \theta_i)\) According to the principle of energy conservation: \[ Q_{lost} = Q_{water} + Q_{calorimeter} \] Substituting the formulas: \[ m_s c_s (\theta_h - \theta_f) = m_w c_w (\theta_f - \theta_i) + m_c c_c (\theta_f - \theta_i) \] We can factor out \((\theta_f - \theta_i)\) on the right side: \[ m_s c_s (\theta_h - \theta_f) = (m_w c_w + m_c c_c)(\theta_f - \theta_i) \] From this equation, we can solve for the specific heat capacity of the solid, \(c_s\): \[ c_s = \frac{(m_w c_w + m_c c_c)(\theta_f - \theta_i)}{m_s (\theta_h - \theta_f)} \] * **Note:** The specific heat capacity of water (\(c_w\)) is approximately \(4200\text{ J/kg}^{\circ}\text{C}\) or \(4.2\text{ J/g}^{\circ}\text{C}\). ### 3. Vocabulary * **Calorimetry:** The scientific study of heat transfer and measurement of heat quantities. * **Heat (Q):** A form of energy transferred between systems or objects with different temperatures. * **Temperature (\(\theta\)):** A measure of the average kinetic energy of the particles in a substance, indicating its degree of hotness or coldness. * **Heat Capacity (C):** The amount of heat energy required to raise the temperature of a specific object by \(1^{\circ}\text{C}\) or \(1\text{ K}\). * **Specific Heat Capacity (c):** The amount of heat energy required to raise the temperature of \(1\text{ kg}\) of a substance by \(1^{\circ}\text{C}\) or \(1\text{ K}\). * **Calorimeter:** An insulated container used to measure heat changes. * **Principle of Conservation of Energy:** Energy cannot be created or destroyed, only transferred from one form or system to another. ### 4. Examples **Example 1: Calculating Heat Energy** How much heat energy is required to raise the temperature of \(2\text{ kg}\) of water from \(20^{\circ}\text{C}\) to \(80^{\circ}\text{C}\)? (Specific heat capacity of water = \(4200\text{ J/kg}^{\circ}\text{C}\)) * **Given:** \(m = 2\text{ kg}\), \(c = 4200\text{ J/kg}^{\circ}\text{C}\), \(\theta_i = 20^{\circ}\text{C}\), \(\theta_f = 80^{\circ}\text{C}\) * **Change in temperature:** \(\Delta\theta = \theta_f - \theta_i = 80^{\circ}\text{C} - 20^{\circ}\text{C} = 60^{\circ}\text{C}\) * **Using the formula:** \(Q = mc\Delta\theta\) \(Q = (2\text{ kg})(4200\text{ J/kg}^{\circ}\text{C}})(60^{\circ}\text{C})\) \(Q = 504000\text{ J}\) or \(504\text{ kJ}\) **Example 2: Applying the Calorimetry Equation** A \(0.5\text{ kg}\) block of an unknown metal at \(100^{\circ}\text{C}\) is placed into a calorimeter containing \(0.8\text{ kg}\) of water at \(20^{\circ}\text{C}\). The calorimeter itself has a mass of \(0.2\text{ kg}\) and is made of aluminum (\(c_{aluminum} = 900\text{ J/kg}^{\circ}\text{C}\)). If the final temperature of the mixture is \(25^{\circ}\text{C}\), calculate the specific heat capacity of the metal. (Specific heat capacity of water = \(4200\text{ J/kg}^{\circ}\text{C}\)) * **Given:** * Solid: \(m_s = 0.5\text{ kg}\), \(\theta_h = 100^{\circ}\text{C}\), \(\theta_f = 25^{\circ}\text{C}\), \(c_s = ?\) * Water: \(m_w = 0.8\text{ kg}\), \(c_w = 4200\text{ J/kg}^{\circ}\text{C}\), \(\theta_i = 20^{\circ}\text{C}\), \(\theta_f = 25^{\circ}\text{C}\) * Calorimeter: \(m_c = 0.2\text{ kg}\), \(c_c = 900\text{ J/kg}^{\circ}\text{C}\), \(\theta_i = 20^{\circ}\text{C}\), \(\theta_f = 25^{\circ}\text{C}\) * **Heat lost by solid:** \(Q_{lost} = m_s c_s (\theta_h - \theta_f) = (0.5) c_s (100 - 25) = 37.5 c_s\) * **Heat gained by water:** \(Q_{water} = m_w c_w (\theta_f - \theta_i) = (0.8)(4200)(25 - 20) = (0.8)(4200)(5) = 16800\text{ J}\) * **Heat gained by calorimeter:** \(Q_{calorimeter} = m_c c_c (\theta_f - \theta_i) = (0.2)(900)(25 - 20) = (0.2)(900)(5) = 900\text{ J}\) * **Conservation of Energy:** \(Q_{lost} = Q_{water} + Q_{calorimeter}\) \(37.5 c_s = 16800 + 900\) \(37.5 c_s = 17700\) \(c_s = \frac{17700}{37.5} = 472\text{ J/kg}^{\circ}\text{C}\) ### 5. Exercises 1. Define heat capacity and specific heat capacity, stating their SI units. 2. What is the difference between heat capacity and specific heat capacity? 3. Calculate the heat energy required to raise the temperature of a \(1.5\text{ kg}\) aluminum block from \(25^{\circ}\text{C}\) to \(75^{\circ}\text{C}\). (Specific heat capacity of aluminum = \(900\text{ J/kg}^{\circ}\text{C}\)). 4. A \(200\text{ g}\) sample of an unknown metal absorbs \(1800\text{ J}\) of heat when its temperature rises by \(30^{\circ}\text{C}\). What is the specific heat capacity of the metal? 5. A kettle contains \(1.8\text{ kg}\) of water at \(20^{\circ}\text{C}\). If the kettle's heating element supplies \(300\text{ kJ}\) of energy, what will be the final temperature of the water? Assume no heat loss to the kettle itself or the surroundings. (Specific heat capacity of water = \(4200\text{ J/kg}^{\circ}\text{C}\)). ### 6. Assignment 1. Explain why water is commonly used as a coolant in car engines and in central heating systems, relating your answer to its specific heat capacity. 2. A copper block of mass \(250\text{ g}\) is heated to \(120^{\circ}\text{C}\) and then transferred quickly into an aluminum calorimeter of mass \(100\text{ g}\) containing \(300\text{ g}\) of water at \(25^{\circ}\text{C}\). The final temperature of the mixture is \(30^{\circ}\text{C}\). * a) Calculate the heat gained by the water. * b) Calculate the heat gained by the calorimeter. * c) Calculate the specific heat capacity of copper. (Given: \(c_{water} = 4200\text{ J/kg}^{\circ}\text{C}\), \(c_{aluminum} = 900\text{ J/kg}^{\circ}\text{C}\)). 3. Discuss the importance of using an insulated calorimeter in experimental determination of specific heat capacity. 4. Derive the formula for specific heat capacity (\(c\)) from the formula for heat energy (\(Q\)). ### 7. Summary In this lesson, we explored the fundamental concepts of calorimetry. We defined **heat capacity (C)** as the heat required to change an object's temperature by \(1^{\circ}\text{C}\), and **specific heat capacity (c)** as the heat required to change the temperature of \(1\text{ kg}\) of a substance by \(1^{\circ}\text{C}\). We learned the key formula for heat energy transfer: \[ Q = mc\Delta\theta \] Finally, we understood how to apply the principle of conservation of energy to experimentally determine the specific heat capacity of a solid using a calorimeter, equating heat lost by the hot solid to heat gained by the water and the calorimeter. These concepts are vital for understanding how energy transfers and affects the temperature of different materials. ## Measurement of specific heat capacity, c for a liquid using Q = mc∆θ and calculations using Q = mc∆θ ### Introduction Welcome to this lesson on calorimetry! In our daily lives, we encounter situations where substances heat up or cool down. Have you ever wondered why some things heat up faster than others, even when given the same amount of heat? This lesson will introduce you to the concept of specific heat capacity, a fundamental property of matter that explains this phenomenon. We will learn how to measure it for a liquid using an experiment and perform calculations involving heat energy transfer. ### Key Concepts #### What is Specific Heat Capacity (c)? Specific heat capacity (often denoted by \( c \)) is a physical property of a substance that quantifies the amount of heat energy required to raise the temperature of a unit mass of that substance by a unit temperature change. In simpler terms, it tells us how much heat a substance can "store" or absorb for a given temperature increase. * **Definition:** The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 kilogram (kg) of the substance by \( 1^ ext{o} ext{C} \) (or 1 Kelvin, K). * **Symbol:** \( c \) * **Units:** Joules per kilogram per degree Celsius (\( ext{J kg}^{-1 ext{o}} ext{C}^{-1} \)) or Joules per kilogram per Kelvin (\( ext{J kg}^{-1} ext{K}^{-1} \)). Since a change of \( 1^ ext{o} ext{C} \) is equal to a change of 1 K, these units are interchangeable for temperature differences. #### The Formula: \( Q = mc heta \) The relationship between heat energy, mass, specific heat capacity, and temperature change is given by a crucial formula: \[ Q = mc\Delta\theta \] Where: * \( Q \) is the amount of heat energy transferred (in Joules, J). * \( m \) is the mass of the substance (in kilograms, kg). * \( c \) is the specific heat capacity of the substance (in \( ext{J kg}^{-1 ext{o}} ext{C}^{-1} \) or \( ext{J kg}^{-1} ext{K}^{-1} \)). * \( \Delta\theta \) (delta theta) is the change in temperature (in \( ^ ext{o} ext{C} \) or K). It is calculated as the final temperature minus the initial temperature (\( \Delta\theta = \theta_{ ext{final}} - \theta_{ ext{initial}} \)). #### Units of Variables It is vital to use consistent units in your calculations: * **Heat energy (\( Q \)):** Joules (J) * **Mass (\( m \)):** Kilograms (kg) * **Specific heat capacity (\( c \)):** \( ext{J kg}^{-1 ext{o}} ext{C}^{-1} \) or \( ext{J kg}^{-1} ext{K}^{-1} \) * **Change in temperature (\( \Delta\theta \)):** \( ^ ext{o} ext{C} \) or K #### Experimental Determination of Specific Heat Capacity (c) of a Liquid We can determine the specific heat capacity of a liquid by supplying a known amount of electrical energy to it and measuring the resulting temperature rise. ##### Apparatus: 1. Immersion heater 2. Liquid (e.g., water, oil) 3. Calorimeter (or beaker, preferably insulated) 4. Thermometer 5. Ammeter 6. Voltmeter 7. Stopwatch 8. Power supply (DC) 9. Electronic balance 10. Stirrer ##### Procedure: 1. Measure the mass of the empty calorimeter, \( m_c \). 2. Pour the liquid into the calorimeter and measure the mass of the calorimeter plus liquid, \( m_t \). Calculate the mass of the liquid, \( m = m_t - m_c \). 3. Set up the apparatus as shown in a typical calorimetry experiment (immersion heater, thermometer, and stirrer immersed in the liquid in the calorimeter, connected to a power supply with ammeter and voltmeter). 4. Measure the initial temperature of the liquid, \( \theta_1 \). 5. Switch on the power supply and immediately start the stopwatch. Record the voltmeter reading (\( V \)) and the ammeter reading (\( I \)). 6. Allow the heater to run for a specific time, \( t \) (e.g., 5-10 minutes), stirring the liquid continuously to ensure uniform temperature. 7. Switch off the power supply and stop the stopwatch. Continue stirring and record the highest final temperature reached, \( \theta_2 \). ##### Calculations: 1. **Electrical energy supplied by the heater (\( Q_{ ext{electrical}} \)):** This is the heat generated by the heater, calculated using the formula for electrical power (\( P = VI \)) multiplied by time (\( t \)). \[ Q_{ ext{electrical}} = P \times t = VIt \] 2. **Heat gained by the liquid (\( Q_{ ext{liquid}} \)):** This is the heat absorbed by the liquid, calculated using the specific heat capacity formula. \[ Q_{ ext{liquid}} = mc\Delta\theta = mc(\theta_2 - \theta_1) \] 3. **Assuming no heat loss to the surroundings:** The electrical energy supplied is entirely converted into heat gained by the liquid. \[ Q_{ ext{electrical}} = Q_{ ext{liquid}} \] \[ VIt = mc(\theta_2 - \theta_1) \] 4. **Rearranging to find \( c \):** \[ c = \frac{VIt}{m(\theta_2 - \theta_1)} \] ##### Precautions to Minimize Heat Loss: * **Insulate the calorimeter:** Use a well-insulated calorimeter (e.g., polystyrene cup, or metal calorimeter placed in a larger container with cotton wool between them) to reduce heat exchange with the surroundings. * **Place a lid on the calorimeter:** To prevent heat loss due to convection and evaporation. * **Stir the liquid continuously:** Ensures uniform temperature throughout the liquid, giving an accurate thermometer reading. * **Minimize heating time:** Reduce the time the experiment runs to lessen the period for heat to escape. * **Take readings carefully:** Ensure accurate measurements of voltage, current, time, mass, and temperature. #### Calculations Using \( Q = mc\Delta\theta \) The formula \( Q = mc\Delta\theta \) is versatile and can be used in various scenarios: * **To find \( Q \):** Given \( m, c, \Delta\theta \). * **To find \( m \):** Given \( Q, c, \Delta\theta \). * **To find \( c \):** Given \( Q, m, \Delta\theta \). * **To find \( \Delta\theta \):** Given \( Q, m, c \). When different substances exchange heat, the principle of conservation of energy applies: **Heat lost by hot objects = Heat gained by cold objects** (assuming no heat loss to the surroundings). ### Vocabulary * **Calorimetry:** The science or act of measuring changes in state variables of a body for the purpose of deriving the heat transfer associated with changes of its state. * **Heat Energy (\( Q \)):** The energy transferred between objects (or systems) due to a temperature difference. * **Specific Heat Capacity (\( c \)):** The amount of heat required to raise the temperature of 1 kg of a substance by \( 1^ ext{o} ext{C} \). * **Temperature (\( \theta \)):** A measure of the average kinetic energy of the particles within a system, indicating its degree of hotness or coldness. * **Immersion Heater:** An electrical device that heats a liquid by direct contact. * **Calorimeter:** A device used for measuring the heat of chemical reactions or physical changes as well as heat capacity. * **Temperature Change (\( \Delta\theta \)):** The difference between the final and initial temperatures of a substance. ### Examples #### Example 1: Calculating Heat Energy (Q) A 2.0 kg block of aluminium (specific heat capacity \( 900 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)) is heated from \( 20^ ext{o} ext{C} \) to \( 80^ ext{o} ext{C} \). Calculate the heat energy absorbed by the block. * **Given:** \( m = 2.0 ext{ kg} \), \( c = 900 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \), \( \theta_{ ext{initial}} = 20^ ext{o} ext{C} \), \( \theta_{ ext{final}} = 80^ ext{o} ext{C} \). * **Find:** \( Q \). * **Solution:** 1. Calculate \( \Delta\theta \): \( \Delta\theta = \theta_{ ext{final}} - \theta_{ ext{initial}} = 80^ ext{o} ext{C} - 20^ ext{o} ext{C} = 60^ ext{o} ext{C} \). 2. Use the formula \( Q = mc\Delta\theta \): \[ Q = (2.0 ext{ kg}) \times (900 ext{ J kg}^{-1 ext{o}} ext{C}^{-1}) \times (60^ ext{o} ext{C}) \] \[ Q = 108,000 ext{ J} \] Therefore, the heat energy absorbed is \( 108,000 ext{ J} \) or \( 108 ext{ kJ} \). #### Example 2: Determining Specific Heat Capacity Experimentally In an experiment to determine the specific heat capacity of an oil, an immersion heater operating at 12 V and 2.0 A was used to heat 0.5 kg of oil for 5 minutes. The temperature of the oil rose from \( 25^ ext{o} ext{C} \) to \( 55^ ext{o} ext{C} \). Calculate the specific heat capacity of the oil, assuming no heat loss. * **Given:** \( V = 12 ext{ V} \), \( I = 2.0 ext{ A} \), \( m = 0.5 ext{ kg} \), \( t = 5 ext{ minutes} \), \( \theta_{ ext{initial}} = 25^ ext{o} ext{C} \), \( \theta_{ ext{final}} = 55^ ext{o} ext{C} \). * **Find:** \( c \). * **Solution:** 1. Convert time to seconds: \( t = 5 ext{ minutes} \times 60 ext{ s/minute} = 300 ext{ s} \). 2. Calculate \( \Delta\theta \): \( \Delta\theta = 55^ ext{o} ext{C} - 25^ ext{o} ext{C} = 30^ ext{o} ext{C} \). 3. Electrical energy supplied (\( Q_{ ext{electrical}} \)): \( Q_{ ext{electrical}} = VIt = (12 ext{ V}) \times (2.0 ext{ A}) \times (300 ext{ s}) = 7200 ext{ J} \). 4. Heat gained by oil (\( Q_{ ext{oil}} \)): \( Q_{ ext{oil}} = mc\Delta\theta = (0.5 ext{ kg}) \times c \times (30^ ext{o} ext{C}) = 15c \). 5. Equate energies (\( Q_{ ext{electrical}} = Q_{ ext{oil}} \)): \( 7200 ext{ J} = 15c \). 6. Solve for \( c \): \[ c = \frac{7200}{15} = 480 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \] The specific heat capacity of the oil is \( 480 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \). #### Example 3: Mixing Liquids 0.8 kg of hot water at \( 80^ ext{o} ext{C} \) is mixed with 0.2 kg of cold water at \( 20^ ext{o} ext{C} \). Assuming no heat loss to the surroundings, what is the final temperature of the mixture? (Specific heat capacity of water = \( 4200 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)). * **Given:** * Hot water: \( m_h = 0.8 ext{ kg} \), \( \theta_{hi} = 80^ ext{o} ext{C} \) * Cold water: \( m_c = 0.2 ext{ kg} \), \( \theta_{ci} = 20^ ext{o} ext{C} \) * \( c_{ ext{water}} = 4200 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \) * **Find:** Final temperature, \( \theta_f \). * **Solution:** 1. Let the final temperature of the mixture be \( \theta_f \). 2. Heat lost by hot water = \( m_h c_{ ext{water}} (\theta_{hi} - \theta_f) \) \( Q_{ ext{lost}} = (0.8)(4200)(80 - \theta_f) \) 3. Heat gained by cold water = \( m_c c_{ ext{water}} (\theta_f - \theta_{ci}) \) \( Q_{ ext{gained}} = (0.2)(4200)(\theta_f - 20) \) 4. Apply the principle of conservation of energy: \( Q_{ ext{lost}} = Q_{ ext{gained}} \) \( (0.8)(4200)(80 - \theta_f) = (0.2)(4200)(\theta_f - 20) \) 5. We can cancel out \( 4200 \) from both sides: \( (0.8)(80 - \theta_f) = (0.2)(\theta_f - 20) \) \( 64 - 0.8\theta_f = 0.2\theta_f - 4 \) \( 64 + 4 = 0.2\theta_f + 0.8\theta_f \) \( 68 = 1.0\theta_f \) \( \theta_f = 68^ ext{o} ext{C} \) The final temperature of the mixture is \( 68^ ext{o} ext{C} \). ### Exercises 1. Calculate the heat energy required to raise the temperature of 1.5 kg of copper (specific heat capacity \( 390 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)) from \( 10^ ext{o} ext{C} \) to \( 70^ ext{o} ext{C} \). 2. A 200 W heater is used to warm 0.4 kg of a liquid for 3 minutes. If the temperature of the liquid increases by \( 45^ ext{o} ext{C} \), calculate the specific heat capacity of the liquid. Assume no heat loss. 3. How much mass of water (specific heat capacity \( 4200 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)) can be heated from \( 20^ ext{o} ext{C} \) to \( 100^ ext{o} ext{C} \) by supplying 336,000 J of heat energy? 4. A metal block of mass 0.5 kg and specific heat capacity \( 400 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \) at \( 100^ ext{o} ext{C} \) is placed into 0.3 kg of water at \( 20^ ext{o} ext{C} \) in an insulated container. What is the final temperature of the mixture? (Specific heat capacity of water = \( 4200 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)). ### Assignment 1. Describe the experimental procedure for determining the specific heat capacity of a liquid using an electrical immersion heater, mentioning the necessary apparatus and precautions to ensure accuracy. 2. An electric kettle rated 2.2 kW contains 1.5 kg of water at \( 25^ ext{o} ext{C} \). If the kettle is 80% efficient, how long will it take to heat the water to its boiling point (\( 100^ ext{o} ext{C} \))? (Specific heat capacity of water = \( 4200 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)). 3. A copper calorimeter of mass 150 g contains 200 g of water at \( 25^ ext{o} ext{C} \). A hot metal sphere of mass 100 g and specific heat capacity \( 380 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \) is dropped into the water. If the final temperature of the mixture is \( 30^ ext{o} ext{C} \), calculate the initial temperature of the metal sphere. (Specific heat capacity of copper = \( 390 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \), specific heat capacity of water = \( 4200 ext{ J kg}^{-1 ext{o}} ext{C}^{-1} \)). ### Summary In this lesson, we explored the concept of specific heat capacity (\( c \)), which is the amount of heat energy required to change the temperature of 1 kg of a substance by \( 1^ ext{o} ext{C} \). We learned the fundamental formula for heat transfer: \[ Q = mc\Delta\theta \] where \( Q \) is heat energy, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta\theta \) is the change in temperature. We also discussed the experimental method to determine the specific heat capacity of a liquid using an immersion heater, emphasizing precautions to minimize heat loss. Finally, we practiced various calculations involving heat energy transfer, including scenarios with electrical heating and mixing of substances, always adhering to the principle of conservation of energy (heat lost = heat gained). Understanding specific heat capacity helps us explain why different materials respond differently to heating and cooling.