10 kg of steam at 100°C with latent heat of vapourisation 2260 kJ is cooled to 50°C. If the specific heat of water is 4200 J/kg°C, find the quantity of heat given out.
Answers
Answered by
0
Answer:
Step-by-step explanation:
Given that,
Mass m=10kg
We know that,
Q
1
=mcΔt
Q
1
=10×4200×(100−50)
Q
1
=2100KJ
Now, the latent heat
Q
2
=mL
Q
2
=10×2260
Q
2
=22600kJ
Now, the total heat energy required
Q=Q
1
+Q
2
Q=2100+22600
Q=24700KJ
Hence, this is the required solution
Answered by
1
Step-by-step explanation:
Given that,
Mass m=10kg
We know that,
Q
1
=mcΔt
Q
1
=10×4200×(100−50)
Q
1
=2100KJ
Now, the latent heat
Q
2
=mL
Q
2
=10×2260
Q
2
=22600kJ
Now, the total heat energy required
Q=Q
1
+Q
2
Q=2100+22600
Q=24700KJ
Hence, this is the required solution
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