Example 5.3: Heat Transfer between Substances at Different Temperatures
Heat Transfer between Substances at Different Temperatures
A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into
425 mL of water at
24.0 °C. The final temperature of the water was measured as
42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (
Table 1), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
\(m_{\mathrm{rebar}}\) \(= 360.\ \mathrm{g}\)
\(V_{\mathrm{water}}\) \(= 425.\ \mathrm{mL}\)
\(T_{\mathrm{\ce{H2O},initial}}\) \(= 24.0\ \mathrm{°aC}\)
\(T_{\mathrm{final}}\) \(= 42.7\ \mathrm{°aC}\)
\(T_{\mathrm{rebar\ inital}}\) = ?
The temperature of the water increases from
24.0 °C to
42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then
heat given off by rebar \( = - \mathrm{heat}\) taken in by water, or:
\(q_{\mathrm{rebar}} = - q_{\mathrm{water}}\)
Since we know how heat is related to other measurable quantities, we have:
\(c_{\mathrm{rebar}} \cdot m_{\mathrm{rebar}} \cdot ΔT_{\mathrm{rebar}} = - c_{\mathrm{water}} \cdot m_{\mathrm{water}} \cdot ΔT_{\mathrm{water}}\)
\(\mathrm{crebar} \cdot \mathrm{mrebar} \cdot (\mathrm{Tf}_{\mathrm{rebar}} - \mathrm{Ti}_{\mathrm{rebar}}) = - \mathrm{cwater} \cdot \mathrm{mwater} \cdot (\mathrm{Tf}_{\mathrm{water}} - \mathrm{Ti}_{\mathrm{water}})\)
Solving for
\(\mathrm{Ti}_{\mathrm{rebar}}\)
\(T_{\mathrm{initial\ rebar}} = \dfrac{c_{\mathrm{water}}}{c_{\mathrm{rebar}}} \cdot \dfrac{m_{\mathrm{water}}}{m_{\mathrm{rebar}}} \cdot (T_{\mathrm{final}} - T_{\mathrm{initial\ water}}) + T_{\mathrm{final}}\)
The density of water is
1.00 g/mL, so
425 mL of water has a mass of
425 g.
\(ρ_{\mathrm{\ce{H2O}}}\) \(= 1.00\ \frac{\mathrm{g}}{\mathrm{mL}}\)
\(m_{\mathrm{water}}\) \(= V_{\mathrm{water}} \cdot ρ_{\mathrm{\ce{H2O}}}\)
\(\ \ \ =425.\ \mathrm{mL} \cdot 1.00\ \frac{\mathrm{g}}{\mathrm{mL}}\)
\(\ \ \ =425.\ \mathrm{g}\)
The specific heats are:
\(c_{\mathrm{water}}\) \(= 4.184\ \frac{\mathrm{J}}{\mathrm{g}\ \mathrm{°ΔC}}\)
\(c_{\mathrm{rebar}}\) \(= 0.449\ \frac{\mathrm{J}}{\mathrm{g}\ \mathrm{°ΔC}}\)
\(\mathrm{Ti}_{\mathrm{rebar}}\) \(= \dfrac{c_{\mathrm{water}}}{c_{\mathrm{rebar}}} \cdot \dfrac{m_{\mathrm{water}}}{m_{\mathrm{rebar}}} \cdot (T_{\mathrm{final}} - T_{\mathrm{\ce{H2O},initial}}) + T_{\mathrm{final}}\)
\(\ \ \ =\dfrac{4.184\ \frac{\mathrm{J}}{\mathrm{g}\ \mathrm{°ΔC}}}{0.449\ \frac{\mathrm{J}}{\mathrm{g}\ \mathrm{°ΔC}}} \cdot \dfrac{425.\ \mathrm{g}}{360.\ \mathrm{g}} \cdot (42.7\ \mathrm{°aC} - 24.0\ \mathrm{°aC}) + 42.7\ \mathrm{°aC}\)
\(\ \ \ =9.318 \cdot 1.181 \cdot 18.70\ \mathrm{°ΔC} + 42.7\ \mathrm{°aC}\)
\(\ \ \ =11.00 \cdot 18.70\ \mathrm{°ΔC} + 42.7\ \mathrm{°aC}\)
\(\ \ \ =205.7\ \mathrm{°ΔC} + 42.7\ \mathrm{°aC}\)
\(\ \ \ =248.\ \mathrm{°aC}\)
The initial temperature of the rebar was
248 °C.