A copper sphere of 100 g mass is heated to raise its temperature to 100^@C and is released in water of mass 195 g and temperature 20^@C in a copper calorimeter. If the mass of calorimeter is 50 g, what will be the maximum temperature of water?

A copper sphere of 100 g mass is heated to raise its temperature to 10

1.	A copper sphere of 100 g mass is heated to raise its temperature to 100^@C and is released in water of mass 195 g and temperature 20^@C in a copper calorimeter. If the mass of calorimeter is 50 g, what will be the maximum temperature of water?
Answer» Solution :Given : Specific heat of copper =0.1 `"cal/g "^@C` Specific heat of calorimeter = 0.1 `"cal/g"^@C` Mass of copper sphere = 100 g Mass of water = 195 g Mass of calorimeter = 50 g Solution: Suppose the copper ball, water and the calorimeter ATTAIN final temperature T. Heat lost by solid OBJECT = heat gained by water in calorimeter + heat gained by calorimeter Here, heat lost by the copper ball = mass of the copper X specific heat of copper x decrease in temperature of ball Q=100 x 0.1 x (100-T) Similarly, Heat gained by the water = mass of the water x specific heat of water x increase in its temperature `Q_1 = 195 xx 1 xx x (T-20)` Heat gained by the calorimeter = mass of the calorimeter x specific heat x increase in its temperature `Q_2` = 50 x 0.1 x (T-20) Now, `Q=Q_1+Q_2` 00 x 0.1 x (100-T) = 195 x 1 x (T-20) + 50 x 0.1 x (T-20) 10(100-T)= 195 (T-20) +5(T-20) 10 (100 - T ) = 200 (T-20) 1000-10T = 200T-4000 210 T = 5000 `T=23.8^@C` The maximum temperature of water will be `23.8^@C`