Question

what will be the final temperature, in °c, of a 373 - g sample of water, initially at 19.1°c, after 72.1 kj have been added to it? note that the specific heat of water is 4.184 j/g°c. report your answer to one place after the decimal point.

Explanation:

Step1: Convert heat from kJ to J

We know that \(1\space kJ = 1000\space J\). So, \(Q=72.1\space kJ=72.1\times1000 = 72100\space J\)

Step2: Use the heat - capacity formula \(Q = mc\Delta T\)

The formula for heat \(Q\) is \(Q=mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). We can re - arrange the formula to find \(\Delta T=\frac{Q}{mc}\). Given \(m = 373\space g\), \(c=4.184\space J/g^{\circ}C\), and \(Q = 72100\space J\). Then \(\Delta T=\frac{72100}{373\times4.184}\)
First, calculate \(373\times4.184=373\times(4 + 0.184)=373\times4+373\times0.184=1492+68.632 = 1560.632\)
Then \(\Delta T=\frac{72100}{1560.632}\approx46.2^{\circ}C\)

Step3: Calculate the final temperature

Since \(\Delta T=T_{final}-T_{initial}\), then \(T_{final}=\Delta T + T_{initial}\). Given \(T_{initial}=19.1^{\circ}C\) and \(\Delta T\approx46.2^{\circ}C\)
\(T_{final}=46.2+19.1=65.3^{\circ}C\)

Answer:

\(65.3^{\circ}C\)