A sphere of aluminium of mass 0-047 kg placed for sufficient time in a vessel containing boiling water- so that the sphere is at 100 -C -It is then immediately transferred to 0-14 kg copper calorimeter containing 0-25 kg of water at 20 -C - The temperature of water rises and attains a steady state at 23 -C -Calculate the specific h capacity of aluminium- Given -specific heat capacity of water is 4186 Jkg - 1K - 1 specific heat capacity of copper calorimeter is 385 Jkg - 1K - 1

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Heat Capacity

A sphere of a...

Question

A sphere of aluminium of mass 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at $100^{\circ} C$ .It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at $20^{\circ} C$ . The temperature of water rises and attains a steady state at $23^{\circ} C$ .Calculate the specific h capacity of aluminium. Given :specific heat capacity of water is 4186 $J k g^{- 1} K^{- 1}$ specific heat capacity of copper calorimeter is 385 $J k g^{- 1} K^{- 1}$

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Solution

Mass of the aluminium sphere ( $m_{1}$ )= 0.047 kg
Initial temperature of the aluminium sphere= $100^{o} C$
Final temperature= $20^{o} C$
Change in temperature ( $Δ T_{1}$ )= $100^{o} C - 2 3^{o} C = 77^{o} C$
Let, specific heat capacity of Al = $c_{1}$
Amount of heat lost by Al= $m 1 \times c 1$ = $Δ T_{1}$
= $0.047 k g \times c_{1} \times 77^{o} c$
Mass of the water $m_{2}$ =0.25 kg
Mass of the calorimeter $m_{3} = 0.14 k g$
Initial temperature of water and calorimeter= $20^{o} C$
Final temperature= $23^{o} C$
Change in temperature ( $Δ T_{2}$ )= $23^{o} C - 20^{o} C = 3^{o} C$
Specific heat capacity of water ( $c_{2}$ )= $4.18 \times 10^{3} J {k g}^{- 1} C^{- 1}$
Total amount of heat gained by calorimeter and water
= $m_{2} \times c_{2} \times Δ T_{2} + m_{3} \times c_{3} \times Δ T_{2}$
= $0.25 \times 4.18 \times 10^{3} \times 3^{o} C + 0.14 \times 0.386 \times 10^{3} \times 3^{o} = 3297.12 J$
At steady state, heat lost by Al= heat gained by water+ heat gained by calorimeter
Therefore $0.047 k g \times c_{1} \times 77^{o} c = 3297.12 J$
$∴ c_{1} = 911.058 J {k g}^{- 1} c^{- 1}$

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