(a) What is meant by the statement: The specific heat capacity of copper is \(400 J kg^{-1}K^{-1}\)? (b)(i) Describe an experiment to determine the specific...
Question 1 Report
(a) What is meant by the statement: The specific heat capacity of copper is \(400 J kg^{-1}K^{-1}\)?
(b)(i) Describe an experiment to determine the specific heat capacity of copper using a copper ball.
(ii) State two precautions necessary to obtain accurate results
(iii) A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = \(4.2 J g^{-1}K^{-1}\)) (Specific heat capacity of copper = \(0.4 Jg^{-1}K^{-1}\)).
Answer Details
(a) "The specific heat capacity of copper is \(400\ \text{J kg}^{-1}\text{K}^{-1}\)" means that 400 joules of heat energy are required to raise the temperature of 1 kilogram of copper by 1 kelvin (1 °C).
(b)(i) Experiment (method of mixtures).
- Weigh the copper ball to get its mass \(m_c\). Heat it in boiling water for some minutes so it reaches a known steady temperature \(\theta_1\) (about 100 °C, read from a thermometer in the water).
- Weigh a calorimeter (mass \(m_{cal}\)), pour in a known mass of water \(m_w\), and record the initial temperature \(\theta_2\) of the calorimeter and water.
- Quickly transfer the hot copper ball into the calorimeter, stir gently, and record the highest (final steady) temperature \(\theta_3\).
- Apply the principle of mixtures: heat lost by copper ball = heat gained by water + heat gained by calorimeter.
Every quantity except the specific heat capacity of copper \(c_c\) is known or measured, so \(c_c\) is calculated.
(b)(ii) Two precautions.
- Transfer the hot ball quickly and lag (insulate) the calorimeter to reduce heat loss to the surroundings.
- Stir the water gently before taking the final temperature so the mixture is uniform.
(b)(iii) Calculation. Let the final steady temperature be \(\theta\).
Heat lost by copper ball \(= 20\times0.4\times(200-\theta)=8(200-\theta)\)
Heat gained by water \(= 50\times4.2\times(\theta-30)=210(\theta-30)\)
Heat gained by calorimeter \(= 60\times0.4\times(\theta-30)=24(\theta-30)\)
\[ 8(200-\theta) = (210+24)(\theta-30) \] \[ 1600 - 8\theta = 234\theta - 7020 \] \[ 8620 = 242\theta \] \[ \theta = 35.6\ ^{\circ}\text{C} \]The final steady temperature of the mixture is about 35.6 °C.
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