\(15\) g of ice at \(0^{\circ}\text{C}\) is added to a vessel containing water at \(40^{\circ}\text{C}\). The mass of water and water equivalent of the vessel is \(60\) g. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be [ \(L_{\text{ice}} = 80\,\text{cal/g}\) , \(S_{\text{water}} = 1\,\text{cal/g}\) ]
Show Hint
First check whether all ice melts, then apply heat lost = heat gained.
Step 1: Use leftover heat
After melting the ice, $2400 - 1200 = 1200$ cal remain. This heats the total mass of $60 + 15 = 75$ g (water equivalent) to the final temperature above $0^\circ\text{C}$.
Step 2: Temperature rise
$\Delta T = \dfrac{1200}{75\times1} = 16^\circ\text{C}$.
Step 3: Result
The mixture settles at $16^\circ\text{C}$.
Step 4: Cross-check
Heat lost by warm water: $60\times(40 - 16) = 1440$ cal. Heat gained: $1200 + 15\times16 = 1440$ cal. The two agree.
Final Answer:
The mixture ends at 16 degrees C. This is option (C).
\[ \boxed{\text{(C) }16^\circ\text{C}} \]
