A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg^@C. It contains 250 gm of liquid at 30^@C having specific heat of 0.4 kca/kg^@C. If we drop a piece of ice of mass 10g at 0^@C, what will be the temperature of the mixture?

A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg^@C. It con

1.	A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg^@C. It contains 250 gm of liquid at 30^@C having specific heat of 0.4 kca/kg^@C. If we drop a piece of ice of mass 10g at 0^@C, what will be the temperature of the mixture?
Answer» Solution :Given : Mass of calorimeter `(m_1)` = 100 g, Specific heat `(c_1) = 0.1 KCAL//kg ^@C`, Temperature `(t_1)=30^@C`, Mass of liquid `(m_2)`=250 g , Specific heat `(c_2) = 0.4 kcal//kg^@ C`, Temperature `(t_2)=30^@C`, Mass of Ice `(m_3)` = 10 g , Specific heat `(c_3) = 0.5 kcal//kg^@C`, Temperature `(t_3)=0^@C` To Find :Final Temperature (T) Solution : Converting GRAMS into calories , 1 g = 7.716 cal 100 g =771.6 cal = 0.7716kcal 250 g = 1929 cal =1.929 kcal 10 g = 77.16 cal =0.07716 kcal Heat capacity of water = 1 kcal/kg Latent heat of fusion of ice = 80 kcal/kg Heat gained or lost= mass X heat capacity x change in temperature Since, ice is put in the liquid, there is heat loss : Change in temperature = (30-T) Heat lost by calorimeter and liquid = heat gained by ice Substituting , [(30-T)0.7716 x 0.1] + [(30-T)1.926 x 0.4 ] = (0.07716 x 80 ) + T x 0.07716 x 1 `therefore ` 2.3148 - 0.07716 T + 23.112 - 0.7704 T = 6.1728 + 0.07716 T `therefore` 0.92472 T = 19.254 `therefore T=20.82^@C` Final temperature of the mixture will be `20.82^@C`