Question

Property	Value
Heat of Fusion (cal/g)	80
Heat of Vaporization (cal/g)	540
Specific Heat (cal/g.${}^o$C)	1.00

For conversion of 10 grams of ice at 0${}^o$ C to liquid water at 100${}^o$C heat needed is:

• A. 1,800 cal
• B. 1,000 cal
• C. 6,200 cal
• D. 800 cal
• E. 7,200 cal

Answer 1

Answer: (A)

1,800 cal

First 10 g ice is melted at 0 deg C.
The heat required is obtained by multiplying mass of water with heat of fusion.
$q_1=10$ g$\times 80$ cal/g$=800$ cal
Next liquid water at 0 deg C is heated to liquid water at 100 deg C.
The heat required is obtained by multiplying mass of water with specific heat and temperature difference.
$q_2=10\times 1.00\times (100-0)=1000$ cal
Total heat required for entire process $=q_1+q_2=800+1000=1800$ cal