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Q.
80 g of water at $30^{\circ} C$ is poured on a large block of ice at $0^{\circ} C$. The mass of ice that melts is
Thermal Properties of Matter
Solution:
Heat lost when water cools from $30^{\circ} C$ to $0^{\circ} C$
$=(80 g )\left(1 \,cal / gm /{ }^{\circ} C \right) \cdot\left(30^{\circ} C \right)=2400 \,cal$
Heat gained by melting ice
$=(\Delta m)\left(80 \frac{\text { cal }}{ gm }\right)$
Heat gained = Heat lost
$80(\Delta m) =2400 $
$\Delta m =\frac{2400}{80}=30\, g$