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Q. $50 g$ ice at $0{ }^{\circ} C$ in kept in an insulating vessel and $50 g$ water at $100{ }^{\circ} C$ is mixed in it. Then the final temperature of the mixture is (neglect the heat loss)

NTA AbhyasNTA Abhyas 2022

Solution:

Heat taken = Heat given
$m_{i}L_{f}+m_{i}s_{w}\left(t - 0\right)=m_{w}s_{w}\left(t_{n} - t\right)$
$50\times 80+50\times 1\times \left(t\right)=50\times 1\times \left(100 - t\right)$
$4000=50\left(100 - 2 t\right)$
$80=\left(100 - 2 t\right)$
$2t=100-80$
$2t=20$
$t=10 \quad{ }^{\circ} C$