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.