Question

- How many calories does it take to heat 10 grams of ice to 6 degrees C?

- How many calories does it take to heat 10 grams of 43 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

Q_{1} = heat required to melt the ice at °C

Q_{1} = ml = (10 g)(334 J/g) = 3340 J

Q_{2} = heat required to raise the temperature of the
liquid water from 0°C to 6°C

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

?T = temperature change = 6° in both K and C scales

Q_{2} = mc?T = (10 g)(4.187 J / gK)(6°K) ? 251 J

Q = Q_{1} + Q_{2} = (3340 + 251) J = 3591 J

1 cal = 4.18 J

Q = (3591 J)[1 cal / (4.18 J)] ? 859 cal

2) Q1 = heat required to warm the water from 43.0 °C to 100.0 °C.

Q2 = heat required to vapourize the water to steam at 100 °C.

Q1=mc?T=10.0 g × 4.187 J°C-1 g-1 ×57.0°C = 2387 J

Q2=m?Hvap=10.0 g × 2260 J?g-1 =226 00 J

Q1+Q2=( 2387 + 226 00) J = 24987 J = 24.98 kJ

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