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Q.
$100\,g$ of ice at $0^{0}C$ is mixed with $100\,g$ of water at $100^\circ C.$ What will be the final temperature of the mixture? (Assume no heat loss)
NTA AbhyasNTA Abhyas 2022
Solution:
$(m L+m s \Delta \theta)_{\text {ice }}=(m s \Delta \theta)_{\text {water }}$
$100(80+t)=100(100-t) \Rightarrow t=10^{\circ} C$.$(m L+m s \Delta \theta)_{\text {ice }}=(m s \Delta \theta)_{\text {water }}$
$100(80+t)=100(100-t) \Rightarrow t=10^{\circ} C$