Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1 °C-1, Latent heat capacity of ice = 330 J g-1.
Given,
mass of water mw = 150 g
mass of calorimeter mc = 50 g
Specific heat capacity of calorimeter = 0.4 J g-1 °C-1,
Specific heat capacity of water = 4.2 J-1 °C-1,
Latent heat capacity of ice = 330 J g-1
mass of ice mi = ?
Heat energy imparted by calorimeter and water contained in it in cooling from 32° C to 5° C is used in melting ice and then raising the temperature of melted ice from 0°C to 5°C.
Heat energy imparted by water (Q1)
= m x c x change in temperature
= 150 x 4.2 x (32 - 5)
= 150 x 4.2 x 27
= 17,010 J
Heat energy imparted by calorimeter (Q2)
= m x c x change in temperature
= 50 x 0.4 x (32 - 5)
= 50 x 0.4 x 27
= 540 J
Heat energy taken by ice to melt (Q3)
= mi x L
= mi x 330
Heat energy gained by water from melted ice to reach from 0°C to 5°C (Q4)
= mi x c x change in temperature
= mi x 4.2 x (5 - 0)
= mi x 4.2 x 5
= mi x 21
From the principle of calorimetry, if the system is fully insulated then,
Heat gained by cold body = Heat lost by hot body.
Q1 + Q2 = Q3 + Q4
Substituting the values we get,
Hence, mass of ice = 50 g