Question

1. suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20°c. what would have been the final temperature of the water?

Explanation:

Step1: Convert units

Convert \(Q = 65\space kJ\) to \(J\): \(Q=65\times10^{3}\space J\), \(m = 450\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\), \(T_{i}=20^{\circ}C\). Use the formula \(Q = mc\Delta T=mc(T_{f}-T_{i})\).

Step2: Solve for \(T_{f}\)

Rearrange the formula for \(T_{f}\): \(T_{f}=\frac{Q}{mc}+T_{i}\). Substitute the values: \(T_{f}=\frac{65\times 10^{3}}{450\times4.18}+20\).
First, calculate \(\frac{65\times 10^{3}}{450\times4.18}=\frac{65000}{1881}\approx 34.56\).
Then \(T_{f}=34.56 + 20\).

Answer:

\(54.56^{\circ}C\)