How much heat energy (in megajoules) is needed to convert 7 kilograms of ice at –9°C to water at 0°C?
A. 2.13 MJ
B. 2.48 MJ
C. 3.09 MJ
D. 3.26 MJ
The correct answer for this question is B. 2.48
Answer: 2471 kJ
Explanation:
The problem can be split into two steps:
1- Calculate the heat energy needed to raise the temperature of the ice from -9 C to 0 C
2- Calculate the heat energy required to melt the ice
Let's solve them separately.
1- The heat energy needed to raise the temperature of the ice from -9 C to 0 C is given by:
[tex]Q=mC_s \Delta T[/tex]
where
m=7 kg is the mass of the ice
Cs=2108 J/kgK is the specific heat capacity of ice
[tex]\Delta T=0 C - (-9 C)=9 C[/tex] is the temperature gain
Substituting into the formula,
[tex]Q_1=(7 kg)(2108 J/kg C)(9 C)=132,804 J=133 kJ[/tex]
2- the heat energy required to melt the ice is given by
[tex]Q = m \lambda_f[/tex]
where
m=7 kg is the mass of the ice
[tex]\lambda_f=334 kJ/kg[/tex] is the specific latent heat of fusion for ice
Substituting into the formula,
[tex]Q_2 = (7 kg)(334 kJ/kg)=2,338 kJ[/tex]
3- So, the total heat energy needed for the entire process is
[tex]Q=Q_1 +Q_2 =133 kJ+2,338 kJ=2,471 kJ[/tex]
