First slide

Questions on calorimetry without phase change

Question

A ball of thermal capacity 10 cal /^o C is heated to the temperature of furnace. It is then transferred into a vessel containing water. The water equivalent of vessel and the contents is 200 gm. The temperature of the vessel and its contents rises from 10^oC to 40^oC. What is the temperature of furnace ?

Moderate

A

640^oC
B

64^oC
C

600^oC
D

100^oC

Solution

Thermal capacity of ball = mc = 10 cal /^o C
Let T be the furnace temperature.
Water eq. of vessel and contents = mc = 200 gm.
Resultant temperature = 40^oC
According to principle of calorimetry,
Heat lost by hot body = heat gained by cold body
$\begin{array}{l} mc (T - 40) = mc (40 - 10) \\ 10 (T - 40) = 200 \times 30 \\ ∴ T = 640^{\circ} C . \end{array}$

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