Question

one litre of water at 1.00°c is warmed by the addition of 4.00 kj of heat energy. what is the final temperature of the water? (1.00 l of water has a mass of 1.00 kg)

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). For water, \(c = 4186\space J/(kg\cdot^{\circ}C)\), \(Q = 4000\space J\), \(m = 1.00\space kg\), and \(T_{initial}=1.00^{\circ}C\).

Step2: Rearrange the formula to solve for \(\Delta T\)

From \(Q = mc\Delta T\), we can get \(\Delta T=\frac{Q}{mc}\). Substitute the values: \(\Delta T=\frac{4000\space J}{1.00\space kg\times4186\space J/(kg\cdot^{\circ}C)}\approx0.956^{\circ}C\).

Step3: Calculate the final temperature \(T_{final}\)

Since \(\Delta T=T_{final}-T_{initial}\), then \(T_{final}=T_{initial}+\Delta T\). Substitute \(T_{initial} = 1.00^{\circ}C\) and \(\Delta T\approx0.956^{\circ}C\). So \(T_{final}=1.00 + 0.956=1.96^{\circ}C\) (rounded to three significant figures).

Answer:

\(1.96^{\circ}C\)