Question

1) How many calories does it take to heat 10 grams of ice to 28 degrees C?

2) How many calories does it take to heat 10 grams of 19 degree C water to a vapor?

Answer 1

1. l = latent heat of fusion of water = 334 kJ/kg = 334 J/g

m = mass of ice = 10 g

Q1 = heat required to melt the ice at °C

Q1 = ml = (10 g)(334 J/g) = 3340 J

Q2 = heat required to raise the temperature of the liquid water from 0°C to 28°C

c = specific heat of liquid water = 4.187 kJ/(kg K) = 4.187 J/(gK)

?T = temperature change = 28° in both K and C scales

Q2 = mc?T = (10 g)(4.187 J / gK)(28°K) ? 1172.36 J

Q = Q1 + Q2 = (3340 + 1172.36) J = 4512.36 J

1 cal = 4.18 J

Q = (4512.36 J)[1 cal / (4.18 J)] ? 1971.41 cal

2. m = mass of ice = 10 g

Lf = 333.7 kJ/kg = 333.7 J/g

You also need the specific heat of water, c.

c = 4.23 kJ/(kg°C) = 4.23 J/(g°C)

The ice melts at a constant temperature of 0°C.  

The heat required to melt it is  Q1 = mLf

After complete melting, additional heat will raise its temperature.

Ti = initial temperature = 0°C

Tf = final temperature = 19°C

The heat required to raise the temperature of the melted ice from Ti to Tf is

Q2 = mc(Tf - Ti)

The total heat required is

Q = Q1 + Q2

Q = mLf + mc(Tf - Ti) = m[Lf + c(Tf - Ti)]

Q = (10 g) [(333.7 J/g) + (4.23 J/(g°C)) (19°C - 0°C)] ? 3417.37 J

You want the answer in calories,

1 cal = 4.18 J

Q = (3417.37 J) / (4.18 J/cal) ? 817.56 cal