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?

Explanation

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 = 0.7716 kcal
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`