Answer to Question #94361 in General Chemistry for India Washington

Question

Suppose that 4535.0 J of heat energy is added to 9 g of ice initially at -8 C. (This could be done, for example, by electrical heating of ice in an insulated container.) What is the final temperature of the water?

Properties of Water:
CP(solid) 1.96 J/g/K
ΔHfus 6.02 kJ/mol
CP(liquid) 4.184 J/g/K

Answer 1

1) Amount of energy spent on heating ice to 0С

$Q_1=C_{p(solid)}*m_{H_2O}*\Delta T=1.96*9*8=141.12J$

2) Amount of energy spent on melting ice

$Q_2=\frac{\Delta H_{fus}*m_{H_2O}*1000}{M_{H_2O}}=\frac{6.02*9*1000}{18}=3010J$

3)Amount of energy available for heating water from 0C

$Q_3=Q-(Q_2+Q_1)=4535.0-(3010+141.12)=1383.88J$

4)Final water temperature

$T=\frac{Q_3}{C_{p(liquid)}*m{H_2O}}=\frac{1383.88}{4.184*9}=38.8C$

Answer: T=38.8C