Example 5.2: Determining Other Quantities
Determining Other Quantities
A piece of unknown metal weighs
348 g. When the metal piece absorbs
6.64 kJ of heat, its temperature increases from
22.4 °C to
43.6 °C. Determine the specific heat of this metal (which might provide a clue to its identity).
Solution
\(m_{\mathrm{unknown}}\) \(= 348.\ \mathrm{g}\)
\(q_{\mathrm{piece}}\) \(= 6.64\ \mathrm{kJ}\)
\(T_{\mathrm{initial}}\) \(= 22.4\ \mathrm{°aC}\)
\(T_{\mathrm{final}}\) \(= 43.6\ \mathrm{°aC}\)
\(c_{\mathrm{unknown}}\) = ?
Since mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship:
\(q = c \cdot m \cdot ΔT = c \cdot m \cdot (T_{\mathrm{final}} - T_{\mathrm{initial}})\)
Solving for c:
\(c = \dfrac{q}{m \cdot ΔT}\)
Substituting the known quantities:
\(c_{\mathrm{unknown}}\) \(= \dfrac{q_{\mathrm{piece}}}{m_{\mathrm{unknown}} \cdot (T_{\mathrm{final}} - T_{\mathrm{initial}})}\)
\(\ \ \ =\dfrac{6.64\ \mathrm{kJ}}{348.\ \mathrm{g} \cdot (43.6\ \mathrm{°aC} - 22.4\ \mathrm{°aC})}\)
\(\ \ \ =\dfrac{6.64\ \mathrm{kJ}}{348.\ \mathrm{g} \cdot 21.20\ \mathrm{°ΔC}}\)
\(\ \ \ =\dfrac{6.64\ \mathrm{kJ}}{7378.\ \mathrm{g}\ \mathrm{°ΔC}}\)
\(\ \ \ =9.00\times 10^{-4}\ \frac{\mathrm{kJ}}{\mathrm{g}\ \mathrm{°ΔC}}\)
\(c_{\mathrm{unknown}}\) \(= 9.00\times 10^{-4}\ \frac{\mathrm{kJ}}{\mathrm{g}\ \mathrm{°ΔC}}\)
\(\ \ \ =900.\ \frac{\mathrm{J}}{\mathrm{kg}\ \mathrm{K}}\)
Comparing this value with the values in
Table 1, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum.