A copper block of mass 3 kg is heated in a furnace to a temperature of 450° C and then placed on a large ice block. Find the maximum amount of ice that can melt? (specific heat of copper = 0.39 Jg^-1K^-1, heat of fusion of water = 335 Jg^-1K^-1)

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 3)

Answer 1

Given:

Mass of copper block (m_Cu) = 3 kg

Temperature of furnace (T_f) = 450° C

Specific heat of copper (s_Cu)= 0.39 Jg-1K-1

Latent heat of fusion of ice (L_f) = 335 Jg-1

Concept:

Heat flows till both the bodies attain the same temperature.

Maximum ice will melt if the heat given by the copper block is only used in melting ice but not raising its temperature.

The temperature of the copper block will rise to furnace temperature

∴ T_Cu = 450° C

Final temperature T_final= 0° C

Formula:

By the principle of calorimetry :

Heat given by hot body = heat gained by cold body

or in this case,

Heat given by copper block = heat gained by ice block

⇒ m_Cus_Cu(T_Cu - T_final) = m_iceL_f

Calculation:

m_Cus_Cu(T_Cu - T_final) = m_iceL_f

⇒ m_ice = m_Cus_Cu(T_Cu - T_final)/L_f

⇒ m_ice = 3 × 0.39 × (450 - 0)/335

⇒ m_ice ≈ 1.57 kg