Chemistry · General chemistry I · Worked example

Calculate the heat along a heating curve

How much heat turns 100.0 g of ice at −10.0 °C into liquid water at 20.0 °C? Use 2.09 J/(g·°C) for ice, 334 J/g to melt ice at 0 °C, and 4.184 J/(g·°C) for liquid water.

q = q_{1} + q_{2} + q_{3}

Split the path at the phase change

The temperature rises to 0 °C, stays there while the ice melts, then rises again. Each segment has its own formula.

The three segments
Segment	What happens	Heat
1	Ice warms from −10.0 °C to 0.0 °C	q = mcΔT
2	Ice melts at 0.0 °C	q = mL
3	Water warms from 0.0 °C to 20.0 °C	q = mcΔT

Warm the ice

ΔT = 0.0 − (−10.0) = 10.0 °C, with the specific heat of ice.

q_{1} = (100.0 g) (2.09 J/ (g \cdot \circ C)) (10.0 \circ C) = 2.09 \times 10^{3} J

\begin{array}{c} q_{1} = (100.0) (2.09) (10.0) J \\ = 2.09 \times 10^{3} J \end{array}

Melt the ice

The temperature stays at 0 °C, so there is no ΔT: each gram takes 334 J.

q_{2} = (100.0 g) (334 J g^{- 1}) = 3.34 \times 10^{4} J

\begin{array}{c} q_{2} = (100.0 g) (334 J g^{- 1}) \\ = 3.34 \times 10^{4} J \end{array}

Warm the water

Now the liquid warms from 0.0 °C to 20.0 °C, with water’s specific heat.

q_{3} = (100.0 g) (4.184 J/ (g \cdot \circ C)) (20.0 \circ C) = 8.37 \times 10^{3} J

\begin{array}{c} q_{3} = (100.0) (4.184) (20.0) J \\ = 8.37 \times 10^{3} J \end{array}

Add the segments

In kilojoules, 2.09 + 33.4 + 8.37 = 43.86, which is 43.9 kJ to the tenths place set by 33.4. Melting alone takes about three quarters of the total.

q = 2.09 + 33.4 + 8.37 = 43.9 kJ

Result

43.9 kJ, of which 33.4 kJ melts the ice.

Your turn

How much heat is released when 50.0 g of liquid water at 30.0 °C cools to 0.0 °C and then freezes completely?

Show the answer and explanation

23.0 kJ is released: q = −23.0 kJ.

Cooling releases (50.0 g)(4.184 J/(g·°C))(30.0 °C) = 6.28 kJ, and freezing releases (50.0 g)(334 J/g) = 16.7 kJ, the reverse of melting. The total is 22.98 kJ, which is 23.0 kJ to the tenths place.

q = - (6.28 kJ + 16.7 kJ) = - 23.0 kJ

\begin{array}{c} q = - (6.28 + 16.7) kJ \\ = - 23.0 kJ \end{array}

Keep exploring

In Calorimetry & heating curves, extend the last segment to 100 °C. Warming the water the rest of the way to its boiling point adds 33.5 kJ, about as much as melting took.

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