Question

0.2 kg of copper heated to 100°c and is dropped into 0.1 kg of water at 30°c contained in a copper calorimeter of mass 0.2 kg if specific heat capacity of copper is 418 j/kg/k and water is 4200 j/kg/k find maximum temperature attained by a water​

Answer 1

Answer:

The answer is 98°C

Explanation:

Let the increasing temperature be T.

Heat lost by copper = mScΔ T

                                  = 0.2 X 4180 X (200  273 - T)

                                  = 836 (473 - T)

Heat gained by water  = mSwΔ T

                                     = 0.1 x 4200 x [ T - ( 30 + 273)]

                                     = 420 [T - 303]

Heat gained by calorimeter  = mScΔ T

                                                = 0.8 x 4180 x [ T - ( 30 + 273)]

                                                =  836 ( T - 303)

Principle of calorimeter:

Heat lost   =  Heat gained

836 (473 - T) =   420 [T - 303] +  836 ( T - 303)

836 (473 - T) =  1256 [ T - 303]

(473 - T)  / ( T - 303) = 1256/836

(473 - T)  / ( T - 303) = 1.5

(473 - T) =  1.5 ( T - 303)

(473 - T ) = 1.5 T - 454 .5

1.5 T + T =  473 +  454.5

2.5 T  = 927.5

T =  927.5/2.5

T = 371 K

T  = 371 - 273  = 98°C