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°C is placed in a Styrofoam cup containing 0.326 kg of water at 20.0°C. The system reaches an equilibrium temperature of 22.4°C. What is the specific heat, [latex]c_x[/latex], of the unknown substance if the heat capacity of the cup is neglected? STRATEGY The water gains thermal energy [latex]Q_{\text {cold }}[/latex] while the block loses thermal energy [latex]Q_{\text {hot }}[/latex]. Using Equation 11.3, substitute expressions into Equation 11.4 and solve for the unknown specific heat, [latex]c_x[/latex]. Q = mcΔT [11.3] [latex]Q_{\mathrm{cold}}=-Q_{\mathrm{hol}}[/latex] [11.4]

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°C is placed in a Styrofoam cup containing 0.326 kg of water at 20.0°C. The system reaches an equilibrium temperature of 22.4°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, substitute expressions into Equation 11.4 and solve for the unknown specific heat, $c_x$ .

Q = mcΔT [11.3]

$Q_{\mathrm{cold}}=-Q_{\mathrm{hol}}$ [11.4]

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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:

$Q_{\text {cold }}=-Q_{\text {hot }}$

$m_w c_w\left(T-T_w\right)=-m_x c_x\left(T-T_x\right)$

Solve for $c_x$ and substitute numerical values:

$c_x=\frac{m_w c_w\left(T – T_w\right)}{m_x\left(T_x – T\right)}$

$=\frac{(0.326 kg )\left(4190 J / kg \cdot{ }^{\circ} C \right)\left(22.4^{\circ} C – 20.0^{\circ} C \right)}{(0.125 kg )\left(90.0^{\circ} C – 22.4^{\circ} C \right)}$

$c_x=388 J / kg \cdot{ }^{\circ} C \rightarrow 390 J / kg \cdot{ }^{\circ} C$

REMARKS Comparing our results to values given in Table 11.1, the unknown substance is probably copper. Note that because the factor (22.4°C – 20.0°C) = 2.4°C has only two significant figures, the final answer must similarly be rounded to two figures, as indicated.

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	0.092 4
Ethyl Alcohol	2 430	0.581
Germanium	322	0.077
Glass	837	0.200
Gold	129	0.030 8
Human tissue	3 470	0.829
Ice	2 090	0.500
Iron	448	0.107
Lead	128	0.030 5
Mercury	138	0.033
Silicon	703	0.168
Silver	234	0.056
Steam	2 010	0.480
Tin	227	0.054 2
Water	4 186	1.00