Question

\(15\) 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
[ \(L_{\text{ice}} = 80\,\text{cal/g}\) , \(S_{\text{water}} = 1\,\text{cal/g}\) ]

• A. \(30^{\circ}\text{C}\)
• B. \(22^{\circ}\text{C}\)
• C. \(16^{\circ}\text{C}\)
• D. \(10^{\circ}\text{C}\)

Hint:

First check whether all ice melts, then apply heat lost = heat gained.

Answer 1

The Correct Option is C

Step 1: Understand the concept
Heat lost by the warm water and vessel goes into melting the ice and then warming the melted ice from \(0^\circ\text{C}\) to the final temperature \(T\).

Step 2: Check whether the ice melts fully
Heat available as the water cools from \(40^\circ\text{C}\) to \(0^\circ\text{C}\): \(60\times1\times40 = 2400\) cal. Heat needed to melt 15 g: \(15\times80 = 1200\) cal. Since \(2400 > 1200\), all the ice melts.

Step 3: Write the heat balance
\[ 60\times1\times(40 - T) = 15\times80 + 15\times1\times(T - 0) \]
\[ 2400 - 60T = 1200 + 15T \]

Step 4: Solve
\(75T = 1200\), so \(T = 16^\circ\text{C}\), option (C). The value \(30^\circ\text{C}\) would result if the latent heat were ignored, and \(10^\circ\text{C}\) is lower than the heat balance allows.

Final Answer:
The final temperature is 16 degrees C. This is option (C). \[ \boxed{\text{(C) }16^\circ\text{C}} \]