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Heat Capacity

A piece of zi...

Question

A piece of zinc at a temperature of $20^{o} C$ weighing 65.38 g is dropped into 180 g of boiling water $(T = 100^{o} C) .$ The specific heat of zinc is $0.4 J g^{- 1 \circ} C^{- 1}$ and that of water is $4.2 J g^{- 1 o} C^{- 1} .$ What is the final common temperature reached by both the zinc and water?

A

97.3^{o} C

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B

33.4^{o} C

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C

80.1^{o} C

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D

60.0^{o} C

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Solution

The correct option is A $97.3^{o} C$
Let final common temperature is $T_{f}$
According to law of energy conservation :
Heat gained by Zinc piece = Heat lost by water
$m_{Z n} s △ T = m_{H_{2} O} s △ T$ , where, s is specific heat capacity.
$65.38 \times 0.4 (T_{f} - 293) = 180 \times 4.2 (373 - T_{f})$
$= \frac{T_{f} - 293}{373 - T_{f}} = 28.90$
$= 29.90 T_{f} = 28.90 \times 373 + 293$
$\Rightarrow T_{f} = 370.32 K$
Hence, final common temperature reached by both the zinc and water $(T_{f}) = {97.3}^{o} C$

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