80 g of water at 30° C is poured on a large block of ice at 0° C. The mass of ice that melts is

Question

80 g of water at 30° C is poured on a large block of ice at 0° C. The mass of ice that melts is (A) 30 g (B) 80 g (C) 150 g (D) 1600 g

Tardigrade Admin · Accepted Answer

Heat lost when water cools from 30° C to 0° C =(80 g )(1 cal / gm / ° C ) ⋅(30° C )=2400 cal Heat gained by melting ice =(Δ m)(80 ( text cal / gm )) Heat gained = Heat lost 80(Δ m) =2400 Δ m =(2400/80)=30 g

Q. 80 g of water at $3 0^{\circ} C$ is poured on a large block of ice at $0^{\circ} C$ . The mass of ice that melts is

30 g

80 g

150 g

1600 g

Heat lost when water cools from $3 0^{\circ} C$ to $0^{\circ} C$
$= (80 g) (1 c a l / g m /^{\circ} C) \cdot (3 0^{\circ} C) = 2400 c a l$
Heat gained by melting ice
$= (Δ m) (80 \frac{cal}{g m})$
Heat gained = Heat lost
$80 (Δ m) = 2400$
$Δ m = \frac{2400}{80} = 30 g$

Q. 80 g of water at 30∘C is poured on a large block of ice at 0∘C. The mass of ice that melts is