A copper sphere of 100 g mass is beated to raise its temperature to `100^@C` g mass is heated to raise its temperature to `100^@C` and is released in water of mass 195 g and teperature `20^@C` in a copper calorimeter. If the mass of clorimeter is 50 g. what will be the maximum temperature of water? (Give : specific heat of copper `= 0.1 cal // g^@C` and specific heat of calrimeter `=0.1 " cal / g "^@C`

Explanation

Let copper sphere, water and calorimeter attain final temperature `T^@C`
We have,

Heat lost by copper sphere
`Q=m_("sphere")xx C_("sphere")xx Delta T`
`100 xx 0.1 xx (100-T)`
Heat gained by water in calorimeter.
`Q_1=m_("water")xxC_("water")xx Delta T`
`=195 xx 1 xx (T-20)`
Heat gained by calorimeter
`Q_2=m_("calorimeter")xxC_("calorimeter")xx Delta T`.
`= 50 xx 0.1xx (T -20)`
According to princilple of heat exchange.
`Q=Q_1+Q_2`
`therefore 10xx(100-T)=195 xx (T-20)+5 xx (T-20)`
`therefore 1000-10T=200(T-20)`
`therefore 210 T= 5000`
`therefore T =23.8^@C`