Example 5.1: Measuring Heat
Measuring Heat
A flask containing
\(8.0\times 10^{2 }\)g of water is heated, and the temperature of the water increases from
21 °C to
85 °C. How much heat did the water absorb?
Solution
\(m_{\mathrm{\ce{H2O}}}\) \(= 8.0\times 10^{2}\ \mathrm{g}\)
\(T_{\mathrm{initial}}\) \(= 21.\ \mathrm{°aC}\)
\(T_{\mathrm{final}}\) \(= 85.\ \mathrm{°aC}\)
\(q_{\mathrm{\ce{H2O}}}\) = ?
To answer this question, consider these factors:
- the specific heat of the substance being heated (in this case, water)
- the amount of substance being heated (in this case,
800 g)
- the magnitude of the temperature change (in this case, from
21 °C to
85 °C).
The specific heat of water is
4.184 J/g °C, so to heat
1 g of water by
1 °C requires
4.184 J. We note that since
4.184 J is required to heat
1 g of water by
1 °C, we will need
800 times as much to heat
800 g of water by
1 °C. Finally, we observe that since
4.184 J are required to heat
1 g of water by
1 °C, we will need
64 times as much to heat it by
64 °C (that is, from
21 °C to
85 °C).
\(c_{\mathrm{\ce{H2O}}}\) \(= 4.184\ \frac{\mathrm{J}}{\mathrm{g}\ \mathrm{°ΔC}}\)
This can be summarized using the equation
\(q = c_{\mathrm{\ce{H2O}}} \cdot m_{\mathrm{\ce{H2O}}} \cdot ΔT\) with
\(ΔT\) \( = T_{\mathrm{final}} - T_{\mathrm{initial}}\)
\(ΔT\) \(= T_{\mathrm{final}} - T_{\mathrm{initial}}\)
\(\ \ \ =85.\ \mathrm{°aC} - 21.\ \mathrm{°aC}\)
\(\ \ \ =64.\ \mathrm{°ΔC}\)
\(q_{\mathrm{\ce{H2O}}}\) \(= c_{\mathrm{\ce{H2O}}} \cdot m_{\mathrm{\ce{H2O}}} \cdot ΔT\)
\(\ \ \ =4.184\ \frac{\mathrm{J}}{\mathrm{g}\ \mathrm{°ΔC}} \cdot 8.0\times 10^{2}\ \mathrm{g} \cdot 64.\ \mathrm{°ΔC}\)
\(\ \ \ =3347.\ \frac{\mathrm{J}}{\mathrm{°ΔC}} \cdot 64.\ \mathrm{°ΔC}\)
\(\ \ \ =2.14\times 10^{5}\ \mathrm{J}\)
\(q_{\mathrm{\ce{H2O}}}\) \(= 2.14\times 10^{5}\ \mathrm{J}\)
\(\ \ \ =214.\ \mathrm{kJ}\)
Because the temperature increased, the water absorbed heat and
\(q_{\mathrm{\ce{H2O}}}\) is positive.