Question

a chemical reaction takes place inside a flask submerged in a water bath. the water bath contains 6.50 kg of water at 28.0 ^ { circ } c. during the reaction 144. kj of heat flows out of the bath and into the flask. calculate the new temperature of the water bath. you can assume the specific heat capacity of water under these conditions is 4.18 j g ^ { - 1 } cdot k ^ { - 1 }. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert units

Convert mass of water from kg to g: \(m = 6.50\space kg=6.50\times10^{3}\space g\). Convert heat from kJ to J: \(q = 144\space kJ = 144\times10^{3}\space J\).

Step2: Use the heat - transfer formula

The heat - transfer formula is \(q=mc\Delta T\), where \(q\) is the heat transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). We want to find \(T_{final}\). Rearranging the formula for \(\Delta T\) gives \(\Delta T=\frac{q}{mc}\).
Substitute \(q=- 144\times10^{3}\space J\) (negative because heat is lost by the water bath), \(m = 6.50\times10^{3}\space g\), and \(c = 4.18\space Jg^{-1}K^{-1}\) into the formula:
\(\Delta T=\frac{-144\times 10^{3}\space J}{6.50\times10^{3}\space g\times4.18\space Jg^{-1}K^{-1}}\)
\(\Delta T=\frac{-144\times10^{3}}{6.50\times4.18\times10^{3}}\space K\)
\(\Delta T=\frac{-144}{6.50\times4.18}\space K\)
\(6.50\times4.18 = 27.17\)
\(\Delta T=\frac{-144}{27.17}\space K\approx - 5.30\space K\)
Since \(\Delta T=T_{final}-T_{initial}\) and \(T_{initial}=28.0^{\circ}C=(28.0 + 273.15)K = 301.15\space K\)
\(T_{final}=T_{initial}+\Delta T\)
\(T_{final}=301.15\space K-5.30\space K = 295.85\space K\)
Convert back to Celsius: \(T_{final}=295.85 - 273.15=22.7^{\circ}C\)

Answer:

\(22.7^{\circ}C\)