Question

$60 \, g$ of ice at $0 \,{}^\circ C$ is added to $200 \, g$ of water initially at $70 \,{}^\circ C$ in a calorimeter of unknown water equivalent $W$ . If the final temperature of the mixture is $40 \,{}^\circ C$ , then the value of $W$ is [Take latent heat of fusion of ice $L_{f}=80 \, cal \, g^{- 1}$ and specific heat capacity of water $s=1 \, cal \, g^{- 1} \,{}^\circ C^{- 1}$ ]

• A. $70g$
• B. $80g$
• C. $40 \, g$
• D. $20g$

Answer 1

Answer: (C)

The principle of calorimetry states that for an isolated system,
The heat gained by the cold body (Here it is ice) = The heat lost by the hot body (Here it is water + calorimeter)
$m_{ice}L_{f}+m_{ice}s\Delta T=\left(m_{water} + W\right)s\Delta T$
$\Rightarrow 60\times 80+60\times 1\times 40=\left(200 + W\right)\times 1\times 30$
$\Rightarrow 480+240=600+3W$
$\Rightarrow W=40g$