Question

what will be the final temperature, in °c, of a 458 - g sample of water, initially at 39.4°c, after 62.8 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 energy unit

We know that \(Q = 62.8\space kJ=62800\space J\), \(m = 458\space g\), \(c = 4.184\space J/g^{\circ}C\), \(T_{initial}=39.4^{\circ}C\). Use the formula \(Q = mc\Delta T\), where \(\Delta T=T_{final}-T_{initial}\).

Step2: Solve for \(\Delta T\)

Rearrange the formula \(\Delta T=\frac{Q}{mc}\). Substitute the values: \(\Delta T=\frac{62800}{458\times4.184}\).
First, calculate \(458\times4.184 = 458\times(4 + 0.184)=458\times4+458\times0.184=1832+84.272 = 1916.272\).
Then \(\Delta T=\frac{62800}{1916.272}\approx32.8\).

Step3: Solve for \(T_{final}\)

Since \(\Delta T=T_{final}-T_{initial}\), then \(T_{final}=\Delta T + T_{initial}\). Substitute \(\Delta T = 32.8\) and \(T_{initial}=39.4\). So \(T_{final}=32.8+39.4 = 72.2\).

Answer:

\(72.2\)