Mixing Ice and Water – Rankers Physics

Calorimetry: Practice Problem & Solution

Assertion (A): If one gram of ice at \(0^{\circ}\text{C}\) is mixed with one gram of water at \(80^{\circ}\text{C}\), then the final temperature of mixture will be \(0^{\circ}\text{C}\). Reason (R): Latent heat of ice is \(540\text{ cal/g}\).  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Calorimetry. Understanding the underlying formula is key to arriving at the correct answer below:

Heat to melt 1g ice at \(0^{\circ}\text{C}\text{ is }1 \times 80\text{ cal/g} = 80\text{ cal}\). Heat from 1g water cooling from \(80^{\circ}\text{C}\text{ to }0^{\circ}\text{C}\text{ is }1 \times 1 \times 80 = 80\text{ cal}\). All ice melts, final temperature is \(0^{\circ}\text{C}\). So (A) is true.

Latent heat of fusion of ice is \(80\text{ cal/g}\), not \(540\text{ cal/g}\). So (R) is false.

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