Question

A calorimeter contains 0.2kg of water at 30 degrees celsius 0.1kg of water at 60 degree celsius is added to it the mixture is well stirred and resulting temperature is found to be 35 degree celsius the thermal capacity of the calorimeter is

Answer 1

Hence the thermal capacity of the calorimeter is 1260 J/K.

EXPLANATION:

By the principle of calorimeter,

Heat gain = heat lost

Het gain  = Thermal capacity of (water + calorimeter) x temp difference

= (0.2 x 4200 + C) (35-30)

where C is the thermal capacity of calorimeter and specific heat of water = 4200

= 5(840 + C)

Similarly, heat lost by water at 60°C  = Thermal capacity of water x temp difference

= 0.1 x 4200 x (60-35)

= 420 x 25

Hence equating both the equation we get,

5(840 + C ) = 420 x 25

=> 840 + C = 2100

=> C = 1260 J/K

Hence the thermal capacity of the calorimeter is 1260 J/K.