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 m_w = 150 g

mass of calorimeter m_c = 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 m_i = ?

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)
= m_i x L
= m_i x 330

Heat energy gained by water from melted ice to reach from 0°C to 5°C (Q₄)
= m_i x c x change in temperature
= m_i x 4.2 x (5 - 0)
= m_i x 4.2 x 5
= m_i 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,

$17010 + 540 = (m$i \times 330) + (mi \times 21) \\[0.5em] 17550 = mi \times 351 \\[0.5em] \Rightarrow mi = \dfrac{17550 }{351} \\[0.5em] \Rightarrow m_i = 50 g

Hence, mass of ice = 50 g