A copper sphere of 100 g mass is heated to raise its temperautre 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 50g,what will the maximum temperature of water?

Explanation

Specific heat of copper = `0.1 cal//g" "^(@)C`
And so specific heat of calorimater = `0.1 cal//g" "^(@)C`
Suppose the copper ball water the calorimeter attain final temperature T.
Heat lost by solid object = heat gained by water in calorimeter + heat gained by the calorimeter.
Here, heat lost by copper ball = mass of the copper `xx` specific heat of copper `xx` decrease in temperature of the ball
`Q = 100 xx 0.1 xx (100-T)`
Similary,
Heat gained by the water = mass of water X its specific heat X increase in its
`Q_(1) = 195 xx 1 xx (T-20)` and
Heat gained by the calorimeter = mass of the calorimeter `xx` its specific heat `xx` increase in its temperature
`Q_(2) = 50 xx 0.1 xx (T-20)`
` Q = Q_(1) + Q_(2)`
`100 xx 0.1 xx (100 -T) = 195 xx 1 xx (T-20) + 50 xx 0.1 xx (T-20)`
`10 (100-T) = 195 (T - 20) +(T-20)`
`210 T = 5000 `
`T = 23.8" "^(@)C`
`therefore` The maximum temperature of water will be `23.8" "^(@)C`.