how much heat is required to convert 10g of ice at -5°c into steam at 100°c 1) Specific heat of ice=2.1 2) Latent heat of vapourisation=2268 3) Latent heat of fusion=336 4) Specific heat of water=4.2
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Answer:
9933 J
Explanation:
Step 1:
Energy required to bring -5° ice to 0° =
E = mcΔt
= 10*2.1*5
= 105 J
Step 2:
Energy required to convert ice to water at 0° =
E = ml
= 10*336
= 3360 J
Step 3:
Energy required to bring water from 0° to 100° =
E = mcΔt
= 10*4.2*100
= 4200 J
Step 4:
Energy required to convert 100° water to 100° steam =
E = ml
= 10*2268
= 22680 J
Step 5:
Total up all the values:
= 105+3360+4200+2268
= 9933 J
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