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Questions on calorimetry without phase change

Question

When 0.400 kg of water at $30^{0} C is mixed with 0.150 kg of water at 25^{0} C contained in a calorimeter,$ $the final temperature$ $is found$
$to be 27^{0} C$ , find the water equivalent of the calorimeter.

Moderate

A

0.450 kg
B

1 kg
C

0.50 kg
D

1.5 kg

Solution

Water of mass $m_{1} = 0.150 kg is taken in the calorimeter at temperature T_{1} = 25^{0} C$ is mixed with N another known mass of pure water $m_{2}$ = 0.400 kg at a temperature $T_{2} = 30^{0} C and final temperature is found to be T = 27^{0} C . Let s_{w}$ and W be the heat capacity of water and the water equivalent of the calorimeter.
Heat gained by (water + calorimeter) at temperature $T_{1} = Heat lost by water at temperature T_{2}$
i.e., $m_{1} s_{w} (T - T_{1}) + {Ws}_{w} (T - T_{1}) = m_{2} s_{w} (T_{2} - T)$
or W = $m_{2} (\frac{T_{2} - T}{T - T_{1}}) - m_{1}$
Hence, W = 0.400 kg [ $\frac{30^{0} C - 27^{0} C}{27^{0} C - 25^{0} C}$ ] - 0.150 kg
= 0.450 kg

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