A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500° C and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat of copper =0.39 J / g ° C heat of fusion of water =335 J / g ).

Question

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500° C and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat of copper =0.39 J / g ° C heat of fusion of water =335 J / g ). (A) 2 kg (B) 1.5 kg (C) 2.5 kg (D) 0.5 kg

Tardigrade Admin · Accepted Answer

Here, mass of copper block, M=2.5 kg =2.5 × 103 g Rise in temperature of the copper block, Δ T=500° C -0° C =500° C Specific heat of copper, c=0.39 J / g ° C Latent heat of fusion of water, L=335 J / g If m is the mass of ice melted, applying the principle of calorimetry, Heat gained by ice = heat lost by copper block i.e., m L=M c Δ T or m=(M c Δ T/L) =(2.5 × 103(g) × 0.39( J / gC °) × 500( C °)/335( J / g )) =1455 g =1.455 ≈ 1.5 kg

Q. A copper block of mass 2.5kg is heated in a furnace to a temperature of 500∘C and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat of copper =0.39J/g∘C heat of fusion of water =335J/g ).

2 kg

1.5 kg

2.5 kg

0.5 kg

Q. A copper block of mass $2.5 k g$ is heated in a furnace to a temperature of $50 0^{\circ} C$ and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat of copper $= 0.39 J / g^{\circ} C$ heat of fusion of water $= 335 J / g$ ).