Question 11.3: Finding a Specific Heat Goal Solve a calorimetry problem inv...

Finding a Specific Heat

Goal Solve a calorimetry problem involving only two substances.

Problem A 125-g block of an unknown substance with a temperature of 90.0^{\circ} \mathrm{C} is placed in a Styrofoam cup containing 0.326 \mathrm{~kg} of water at 20.0^{\circ} \mathrm{C}. The system reaches an equilibrium temperature of 22.4^{\circ} \mathrm{C}. What is the specific heat, c_{x}, of the unknown substance if the heat capacity of the cup is neglected?

Strategy The water gains thermal energy Q_{\text {cold }}, while the block loses thermal energy Q_{\text {hot. }} Using Equation 11.3,

Q=m c\Delta T        (11.3)

substitute expressions into Equation 11.4

Q_{\mathrm{cold}}=-Q_{\mathrm{hol}}      (11.4)

and solve for the unknown specific heat, c_{x}.

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Let T be the final temperature, and let T_{w} and T_{x} be the initial temperatures of the water and block, respectively. Apply Equations 11.3 and 11.4:

\begin{aligned} Q_{\text {cold }} & =-Q_{\text {hot }} \\ m_{w} c_{w}\left(T-T_{w}\right) & =-m_{x} c_{x}\left(T-T_{x}\right) \end{aligned}

Solve for c_{x} and substitute numerical values:

\begin{aligned} c_{x} & =\frac{m_{w} c_{w}\left(T-T_{w}\right)}{m_{x}\left(T_{x}-T\right)} \\ & =\frac{(0.326 \mathrm{~kg})\left(4190 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(22.4^{\circ} \mathrm{C}-20.0^{\circ} \mathrm{C}\right)}{(0.125 \mathrm{~kg})\left(90.0^{\circ} \mathrm{C}-22.4^{\circ} \mathrm{C}\right)} \\ c_{x} & =388 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C} \end{aligned}

Remarks Comparing our results to values given in Table 11.1, the unknown substance is probably copper.

TABLE 11.1
Specific Heats of Some Materials at Atmospheric Pressure

Substance J/kg · °C cal/g · °C
Aluminum 900 0.215
Beryllium 1,820 0.436
Cadmium 230 0.055
Copper 387 0.0924
Germanium 322 0.077
Glass 837 0.200
Gold 129 0.0308
Ice 2,090 0.500
Iron 448 0.107
Lead 128 0.0305
Mercury 138 0.033
Silicon 703 0.168
Silver 234 0.056
Steam 2,010 0.480
Water 4,186 1.00

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