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InterviewSolution
| 1. |
540g of ice at 0°C is mixed with 540g of water at 80°C. The temperature of mixture is : (a) 40°C (b) 80°C (c) 0°C (d) less than 0°C |
| Answer» <p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000"><strong>Correct option (c) 0°C </strong></span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000"><strong>Explanation:</strong></span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">The amount of heat required to melt 540g of ice is given by </span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= 540 × 80 = 43200 cal </span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">(latent heat of ice = 80 kcal ⁄ g) </span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Now heat lost by 540 g of water form 80°C to 0°C.</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= 540 × 1 × (80° − 0°)</span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">= 43200 cal </span></span></p><p style="text-align: justify;"><span style="font-family:Arial,Helvetica,sans-serif"><span style="color:#000000">Therefore, all the ice must be melt and temperature of mixture is zero.</span></span></p> | |