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Some constants which you may find useful Specific heat capacity of water=4200 jkg^-1K^-1 Latent heat of fusion of ice=334J/g Radius of thc sun=6.96times 10^8m Specific heat capacity of copper=390Jkg^-1K^-1 Molar gas constant R=8.31JK^-1Kmol^-1 QUESTION ONE (30 MARKS) a) Briefly define the following terms (i) Internal energy (1 Mark) (ii) Heat (1 Mark) b) State the Zeroth Law of thermodynamics (1 Mark) c) When do we say two bodies are in thermal equilibrium? (1 Mark) d) Define the following terms: (i) Thermal capacity (1 Mark) (ii) Specific heat capacity (1 Mark) e) A metal of mass 0.2 kg at 100^circ C is dropped into 0.8 kg of water at 15^circ C contained in a container of mass 012 kg and specific heat capacity=400Jkg^-1K^-1 . The final temperature reached is 35^circ C Assuming no heat losses calculate the specific heat capacity of the metal. (5 Marks)

Вопрос

Some constants which you may find useful
Specific heat capacity of water=4200 jkg^-1K^-1
Latent heat of fusion of ice=334J/g
Radius of thc sun=6.96times 10^8m
Specific heat capacity of copper=390Jkg^-1K^-1
Molar gas constant R=8.31JK^-1Kmol^-1
QUESTION ONE (30 MARKS)
a) Briefly define the following terms
(i) Internal energy
(1 Mark)
(ii) Heat
(1 Mark)
b) State the Zeroth Law of thermodynamics
(1 Mark)
c) When do we say two bodies are in thermal equilibrium?
(1 Mark)
d) Define the following terms:
(i) Thermal capacity
(1 Mark)
(ii) Specific heat capacity
(1 Mark)
e) A metal of mass 0.2 kg at 100^circ C is dropped into 0.8 kg of water at 15^circ C contained in a
container of mass 012 kg and specific heat capacity=400Jkg^-1K^-1 . The final
temperature reached is 35^circ C Assuming no heat losses calculate the specific heat capacity
of the metal.
(5 Marks)

Some constants which you may find useful Specific heat capacity of water=4200 jkg^-1K^-1 Latent heat of fusion of ice=334J/g Radius of thc sun=6.96times 10^8m Specific heat capacity of copper=390Jkg^-1K^-1 Molar gas constant R=8.31JK^-1Kmol^-1 QUESTION ONE (30 MARKS) a) Briefly define the following terms (i) Internal energy (1 Mark) (ii) Heat (1 Mark) b) State the Zeroth Law of thermodynamics (1 Mark) c) When do we say two bodies are in thermal equilibrium? (1 Mark) d) Define the following terms: (i) Thermal capacity (1 Mark) (ii) Specific heat capacity (1 Mark) e) A metal of mass 0.2 kg at 100^circ C is dropped into 0.8 kg of water at 15^circ C contained in a container of mass 012 kg and specific heat capacity=400Jkg^-1K^-1 . The final temperature reached is 35^circ C Assuming no heat losses calculate the specific heat capacity of the metal. (5 Marks)

Решения

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Христина
Экспертная проверкаЭкспертная проверка
профессионал · Репетитор 6 лет

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a) Definitions:<br />(i) Internal energy: The total energy possessed by a system due to the kinetic and potential energies of its molecules. It includes the energy associated with the random motion of particles and the energy associated with the interactions between particles.<br /><br />(ii) Heat: The transfer of energy from one body to another due to a temperature difference between them. It is the energy that flows from a hotter body to a cooler body.<br /><br />b) Zeroth Law of Thermodynamics: If two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.<br /><br />c) Two bodies are said to be in thermal equilibrium when they have the same temperature. This means that there is no net transfer of heat between them.<br /><br />d) Definitions:<br />(i) Thermal capacity: The amount of heat required to raise the temperature of a body by one degree Celsius. It is a measure of a body's ability to absorb heat.<br /><br />(ii) Specific heat capacity: The amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a measure of a substance's ability to absorb heat per unit mass.<br /><br />e) To calculate the specific heat capacity of the metal, we can use the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the water and the container.<br /><br />Heat lost by the metal = Heat gained by the water + Heat gained by the container<br /><br />Using the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can set up the equation:<br /><br />m_metal * c_metal * (T_initial_metal - T_final) = m_water * c_water * (T_final - T_initial_water) + m_container * c_container * (T_final - T_initial_container)<br /><br />Substituting the given values:<br /><br />0.2 kg * c_metal * (100°C - 35°C) = 0.8 kg * 4200 Jkg^-1K^-1 * (35°C - 15°C) + 0.12 kg * 400 Jkg^-1K^-1 * (35°C - 15°C)<br /><br />Solving for c_metal:<br /><br />c_metal = (0.8 kg * 4200 Jkg^-1K^-1 * 20°C + 0.12 kg * 400 Jkg^-1K^-1 * 20°C) / (0.2 kg * 65°C)<br /><br />c_metal = (67200 J + 960 J) / 13 J<br /><br />c_metal = 51720 J / 13 J<br /><br />c_metal ≈ 3985.38 Jkg^-1K^-1<br /><br />Therefore, the specific heat capacity of the metal is approximately 3985.38 Jkg^-1K^-1.
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