Question

Please help me in this sum below

A calorie-meter contains 0.57 kg of water at 30 degree Celsius. 0.2 kg of water at 50 degree Celsius is thoroughly mixed with it. The final temperature of the mixture becomes 35 degree Celsius. Find the water equivalent of the calorie-meter.

Answer 1

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.