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Question

Calculate the heat required to convert  3kg of ice at12C kept in a calorimeter to steam at 100 at atmospheric pressure. Given : specific heat of ice =2100×105Jkg1K1 , specific heat of water =4186×103Jkg1K1 , latent heat of fusion of ce=3.35×105Jkg1 and latent heat of steam =2.256×106Jkg1.

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Solution

Step 1: Bringing ice at 0° C
Q1 = MCΔT
=>3*2100*(12-0)
=> 75600KgJ
Step 2: Changin to water at 0°C
Q2=ML
=>3*3.35*10
=>100.5Jkg
Step 3: Bringing water to 100°C
Q3=MCΔT
=>3*4.186*100
=>1255.8JKg
Step4: Converting water to steam
Q4=ML
=>3*22.576JKg
=>67.728JKg
Total heat == Q1+Q2+Q3+Q4
==> 77024.03J

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