Physics · Heat, temperature, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat
Bunty mixed 440 gm of ice at with of water at in a bowl. Then what would remain
Bunty mixed 440 gm of ice at \( 0^{\circ} \mathrm{C} \) with \( 540 \mathrm{gm} \) of water at \( 80^{\circ} \mathrm{C} \) in a bowl. Then what would remain after sometime in the bowl?
- A. only ice
- B. only water
- C. ice and water in same amount
- D. ice and water will vapourise
Step-by-step solution
The heat required to melt all ice is 440 g × 80 cal/g = 35200 cal. The heat released by cooling water from 80°C to 0°C is 540 g × 1 cal/g°C × 80°C = 43200 cal. Since the available heat (43200 cal) exceeds the heat needed to melt ice (35200 cal), all ice melts, leaving only water at a final temperature above 0°C.
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