First slide

Change of state

Question

80 gm of water at 30^oC is poured on a large block of ice at 0^oC. The mass of ice that melts is

Moderate

A

30 gm
B

80 gm
C

150 gm
D

1600 gm

Solution

Since, the block of ice at 0^oC is large, the whole of ice will not melt, hence final temperature is 0^oC .
$∴$ Q₁ = heat given by water in cooling upto 0^oC
$= msΔθ = 80 \times 1 \times (30 - 0)$
= 2400 cal
If m gm be the mass of ice melted, then
$Q_{2} = {mL}_{F} = m \times 80 Now, Q_{2} = Q_{1}$
m x 80 = 2400 or m = 30 gm.

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