A copper block of mass 50 g is heated to 100° C and placed on a block of ice at 0° C. The specific heat of copper is 0.1 cal g -1 ° C -1 and latent heat of ice is 80 cal g -1 . The amount of ice melted is

Question

A copper block of mass 50 g is heated to 100° C and placed on a block of ice at 0° C. The specific heat of copper is 0.1 cal g -1 ° C -1 and latent heat of ice is 80 cal g -1 . The amount of ice melted is (A) 6.15 g (B) 6.2 g (C) 6.25 g (D) 6.3 g

Tardigrade Admin · Accepted Answer

Heat lost by copper = Heat gained by ice (50 g )(0.1. cal . g -1 C -1)(100-0)° C =(Δ m) ⋅ 80 ⇒ Δ m=(50 × 0.1 × 100/80)=6.25 g

Q. A copper block of mass 50 g is heated to $10 0^{\circ} C$ and placed on a block of ice at $0^{\circ} C$ . The specific heat of copper is 0.1 cal $g^{- 1}^{\circ} C^{- 1}$ and latent heat of ice is 80 cal $g^{- 1} .$ The amount of ice melted is

6.15 g

6.2 g

6.25 g

6.3 g

Heat lost by copper = Heat gained by ice
$(50 g) (0.1$ cal $g^{- 1} C^{- 1}) (100 - 0)^{\circ} C = (Δ m) \cdot 80$
$\Rightarrow Δ m = \frac{50 \times 0.1 \times 100}{80} = 6.25 g$

Q. A copper block of mass 50 g is heated to 100∘C and placed on a block of ice at 0∘C. The specific heat of copper is 0.1 cal g−1∘C−1 and latent heat of ice is 80 cal g−1. The amount of ice melted is