Question

what will be the final temperature, in °c, of a 524 - g sample of water, initially at 31.3 °c, after 72.8 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, \(q=72.8\space kJ\times1000 = 72800\space J\)

Step2: Use the heat - transfer formula \(q = mc\Delta T\)

The formula for heat transfer is \(q=mc\Delta T\), where \(q\) is the heat added, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). We can re - arrange the formula to solve for \(\Delta T\): \(\Delta T=\frac{q}{mc}\)
Given \(m = 524\space g\), \(c=4.184\space J/g^{\circ}C\), and \(q = 72800\space J\)
\(\Delta T=\frac{72800\space J}{524\space g\times4.184\space J/g^{\circ}C}\)
First, calculate the denominator: \(524\times4.184 = 524\times(4 + 0.184)=524\times4+524\times0.184=2096+96.416 = 2192.416\)
Then, \(\Delta T=\frac{72800}{2192.416}\approx33.2\space^{\circ}C\)

Step3: Calculate the final temperature

Since \(\Delta T=T_{final}-T_{initial}\), then \(T_{final}=\Delta T + T_{initial}\)
Given \(T_{initial}=31.3\space^{\circ}C\)
\(T_{final}=33.2+31.3 = 64.5\space^{\circ}C\)

Answer:

\(64.5\space^{\circ}C\)