Question

1. 250 kj of energy are removed from a 4.00 x 10² g sample of water at 60°c. will the sample of water completely freeze? explain.

Explanation:

Step1: Calculate energy to cool water to \(0^{\circ}C\)

Use \(Q = mc\Delta T\). \(m = 400\space g\), \(c = 4.18\space J/(g\cdot^{\circ}C)\), \(\Delta T=60^{\circ}C\).
\(Q_1=400\times4.18\times60 = 100320\space J = 100.32\space kJ\)

Step2: Calculate energy to freeze water

Use \(Q = m\Delta H_{fus}\). \(\Delta H_{fus}=334\space J/g\), \(m = 400\space g\).
\(Q_2=400\times334 = 133600\space J=133.6\space kJ\)

Step3: Calculate total energy

\(Q_{total}=Q_1 + Q_2=100.32 + 133.6 = 233.92\space kJ\)

Answer:

Since \(Q_{total}=233.92\space kJ<250\space kJ\), the sample of water will completely freeze.