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WAEC Physics 2011 Paper

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Question : 21

Total: 49

A metal of mass 200g at a temperature of 100^oC is placed in 100g of water at 25^oC in a container of negligible heat capacity. If the final steady temperature is 30^oC, calculate the specific heat capacity of the metal. [specific heat capacity of the water is 4200 Jkg^-1 K^-1]

150 Jkg^-1K^-1
300 Jkg^-1K^-1
320 Jkg^-1K^-1
1960 Jkg^-1K^-1

Solution:

Concept:

Heat lost by the hot metal equals heat gained by the cooler water, because the container has negligible heat capacity.

Explanation:

Mass of metal,

m_m = 200 \, \text{g} = 0.2 \, \text{kg}

.

Temperature drop of metal,

\Delta T_m = 100 \, {}^{\circ}\text{C} - 30 \, {}^{\circ}\text{C} = 70 \, \text{K}

.

Mass of water,

m_w = 100 \, \text{g} = 0.1 \, \text{kg}

.

Temperature rise of water,

\Delta T_w = 30 \, {}^{\circ}\text{C} - 25 \, {}^{\circ}\text{C} = 5 \, \text{K}

.

Specific heat capacity of water,

c_w = 4200 \, \text{J kg}^{-1}\text{K}^{-1}

.

Let the specific heat capacity of the metal be

c_m

.

Using heat lost = heat gained:

m_m c_m \Delta T_m = m_w c_w \Delta T_w

0.2 \times c_m \times 70 = 0.1 \times 4200 \times 5

14 \, c_m = 2100

c_m = \frac{2100}{14} = 150 \, \text{J kg}^{-1}\text{K}^{-1}

Answer:

A.

150 \, \text{J kg}^{-1}\text{K}^{-1}

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