Question:
\(540\) gram of ice at \(0^{\circ }\)C is mixed with \(540\) gram of water at \(90^{\circ }\)C. The final temperature of the mixture will be [Latent heat of fusion of water = 80 cal/g]
\(540\) gram of ice at \(0^{\circ }\)C is mixed with \(540\) gram of water at \(90^{\circ }\)C. The final temperature of the mixture will be [Latent heat of fusion of water = 80 cal/g]
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Check first whether all the ice can melt using the heat from the hot water.
Updated On: Nov 12, 2026
- \(10^{\circ }\)C
- \(15^{\circ }\)C
- \(5^{\circ }\)C
- \(30^{\circ }\)C
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Heat lost by hot water equals heat gained by ice. First check if all ice melts.
Step 2: Heat balance check
Heat to melt ice \(=540\times80=43200\) cal. Heat released by water cooling to \(0^{\circ}\)C \(=540\times90=48600\) cal. All ice melts, and 5400 cal remains.
Step 3: Final temperature
Total mass \(=1080\) g, so \(T=\dfrac{5400}{1080}=5^{\circ}\)C.
Final Answer:
The final temperature of the mixture is 5 degrees Celsius. \[ \boxed{\text{(C) }5^{\circ}\text{C}} \]
Heat lost by hot water equals heat gained by ice. First check if all ice melts.
Step 2: Heat balance check
Heat to melt ice \(=540\times80=43200\) cal. Heat released by water cooling to \(0^{\circ}\)C \(=540\times90=48600\) cal. All ice melts, and 5400 cal remains.
Step 3: Final temperature
Total mass \(=1080\) g, so \(T=\dfrac{5400}{1080}=5^{\circ}\)C.
Final Answer:
The final temperature of the mixture is 5 degrees Celsius. \[ \boxed{\text{(C) }5^{\circ}\text{C}} \]
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