Question

what will be the final temperature, in °c, of a 421 - g sample of water, initially at 19.8°c, after 65.7 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, \(q=65.7\space kJ=65.7\times1000 = 65700\space J\)

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 \(m = 421\space g\), \(c = 4.184\space J/g^{\circ}C\), and \(q = 65700\space J\)
\(\Delta T=\frac{65700}{421\times4.184}\)
First, calculate \(421\times4.184=421\times(4 + 0.184)=421\times4+421\times0.184=1684+77.464 = 1761.464\)
Then \(\Delta T=\frac{65700}{1761.464}\approx37.3^{\circ}C\)

Step3: Calculate the final temperature

Since \(\Delta T=T_{final}-T_{initial}\), then \(T_{final}=\Delta T + T_{initial}\)
Given \(T_{initial}=19.8^{\circ}C\)
\(T_{final}=37.3+19.8 = 57.1^{\circ}C\)

Answer:

\(57.1^{\circ}C\)