Question:
If $Q_1$ is heat to convert 2 g ice at 0°C to water at 40°C and $Q_2$ for 4 g ice at 0°C to water at 80°C, find $Q_1:Q_2$ (L = 80 cal/g, c = 1 cal/g°C):
If $Q_1$ is heat to convert 2 g ice at 0°C to water at 40°C and $Q_2$ for 4 g ice at 0°C to water at 80°C, find $Q_1:Q_2$ (L = 80 cal/g, c = 1 cal/g°C):
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Always separate latent heat and sensible heat.
Updated On: Jun 17, 2026
- 3:8
- 1:2
- 1:4
- 3:4
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The Correct Option is A
Solution and Explanation
Step 1: For $Q_1$: \[ Q_1 = mL + mc\Delta T \] \[ = 2(80) + 2(1)(40) = 160 + 80 = 240 \]
Step 2: For $Q_2$: \[ Q_2 = 4(80) + 4(1)(80) = 320 + 320 = 640 \]
Step 3: Ratio: \[ Q_1 : Q_2 = 240 : 640 = 3 : 8 \]
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