Question

1. an ice cube tray full of ice (235g) at -7.0°c is allowed to warm up to room temperature (22°c). how much energy must be absorbed by the contents of the tray in order for this to happen?

Explanation:

Step1: Calculate energy for ice heating

Use formula \(Q = mc\Delta T\). For ice, \(c_{ice}=2.09\,\text{J/g}\cdot^{\circ}\text{C}\), \(m = 235\,\text{g}\), \(\Delta T=T_2 - T_1=0 - (-7)=7^{\circ}\text{C}\).
\(Q_1=235\times2.09\times7\)
\(Q_1 = 235\times14.63=3438.05\,\text{J}\)

Step2: Calculate energy for ice - water phase change

Use formula \(Q = mL_f\). \(L_f = 334\,\text{J/g}\), \(m = 235\,\text{g}\)
\(Q_2=235\times334 = 78490\,\text{J}\)

Step3: Calculate energy for water heating

For water, \(c_{water}=4.18\,\text{J/g}\cdot^{\circ}\text{C}\), \(m = 235\,\text{g}\), \(\Delta T=22 - 0=22^{\circ}\text{C}\)
\(Q_3=235\times4.18\times22\)
\(Q_3=235\times91.96 = 21610.6\,\text{J}\)

Step4: Calculate total energy

\(Q_{total}=Q_1 + Q_2+Q_3\)
\(Q_{total}=3438.05+78490 + 21610.6\)
\(Q_{total}=103538.65\,\text{J}\approx1.04\times 10^{5}\,\text{J}\)

Answer:

The energy absorbed by the contents of the tray is approximately \(1.04\times 10^{5}\,\text{J}\)