Chemistry · General chemistry I · Worked example

Find the specific heat of a metal

A 150.0 g metal block at 100.0 °C is dropped into 100.0 g of water at 20.0 °C in an insulated cup, and both end at 31.1 °C. What is the metal’s specific heat? Ignore the heat taken up by the cup.

q_{metal} + q_{water} = 0

Find each temperature change

Each ΔT is final minus initial. The water warms by 11.1 °C. The metal cools by 68.9 °C, so its ΔT is negative.

\begin{array}{c} Δ T_{water} = 31.1 - 20.0 = 11.1 \circ C \\ Δ T_{metal} = 31.1 - 100.0 = - 68.9 \circ C \end{array}

Find the heat the water gained

Use q = mcΔT with water’s specific heat, 4.184 J/(g·°C). The water absorbs 4.64 × 10³ J.

q_{water} = (100.0 g) (4.184 J/ (g \cdot \circ C)) (11.1 \circ C) = 4.64 \times 10^{3} J

\begin{array}{c} q_{water} = (100.0) (4.184) (11.1) J \\ = 4.64 \times 10^{3} J \end{array}

Balance the heat

No heat escapes, so the metal released exactly what the water absorbed.

q_{metal} = - q_{water} = - 4.64 \times 10^{3} J

\begin{array}{c} q_{metal} = - q_{water} \\ = - 4.64 \times 10^{3} J \end{array}

Solve for the specific heat

Rearrange q = mcΔT for c and use the metal’s own mass and ΔT. The two negative signs cancel, as they must: a specific heat is positive.

c = \frac{- 4.64 \times 10^{3} J}{(150.0 g) (- 68.9 \circ C)} = 0.449 J/ (g \cdot \circ C)

\begin{array}{c} c = \frac{- 4.64 \times 10^{3}}{(150.0) (- 68.9)} \\ = 0.449 J/ (g \cdot \circ C) \end{array}

Check the result

With c = 0.449 J/(g·°C), the heat balance puts the final temperature at 31.1 °C, as measured. The metal’s specific heat is about a tenth of water’s, which is why the metal cooled by 68.9 °C while the water warmed by only 11.1 °C.

Result

The metal’s specific heat is 0.449 J/(g·°C).

Your turn

A 200.0 g metal cylinder at 100.0 °C is placed in 50.0 g of water at 20.0 °C, and both end at 41.6 °C. Find the metal’s specific heat.

Show the answer and explanation

0.387 J/(g·°C).

The water gains (50.0 g)(4.184 J/(g·°C))(21.6 °C) = 4.52 × 10³ J, so the metal loses 4.52 × 10³ J while its temperature changes by −58.4 °C. Then c = −4.52 × 10³ J ÷ (200.0 g × −58.4 °C) = 0.387 J/(g·°C).

c = \frac{- 4.52 \times 10^{3} J}{(200.0 g) (- 58.4 \circ C)} = 0.387 J/ (g \cdot \circ C)

\begin{array}{c} c = \frac{- 4.52 \times 10^{3}}{(200.0) (- 58.4)} \\ = 0.387 J/ (g \cdot \circ C) \end{array}

Keep exploring

In Calorimetry & heating curves, double the water to 200.0 g. The final temperature falls to 26.0 °C, because the same metal now warms twice as much water.

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