Question

unit 3 worksheet 4 - quantitative energy problems
part 2
energy constants (h₂o)
334 j/g heat of fusion (melting or freezing) hf
2260 j/g heat of vaporization (evaporating or condensing) hy
2.1 j/g°c heat capacity (c) of solid water
4.18 j/g°c heat capacity (c) of liquid water
for each of the problems sketch a warming or cooling curve to help you decide which equation(s) to use to solve the problem. keep a reasonable number of sig figs in your answers.

1. how much energy must be absorbed by a 150 g sample of ice at 0.0°c that melts and then warms to 25.0°c?
2. suppose in the icy hot lab that the burner transfers 325 kj of energy to 450 g of liquid water at 20.°c. what mass of the water would be boiled away?
3. a 12oz can of soft drink (assume m = 340 g) at 25°c is placed in a freezer where the temperature is - 12°c. how much energy must be removed from the soft drink for it to reach this temperature?

Explanation:

1.

Step1: Calculate heat for melting

Use formula \(Q = mH_f\). Here \(m = 150g\) and \(H_f=334J/g\).
\(Q_1=150g\times334J/g = 50100J\)

Step2: Calculate heat for warming

Use formula \(Q = mc\Delta T\). Here \(m = 150g\), \(c = 4.18J/g^{\circ}C\), \(\Delta T=(25 - 0)^{\circ}C\)
\(Q_2=150g\times4.18J/g^{\circ}C\times25^{\circ}C=15675J\)

Step3: Calculate total heat

\(Q_{total}=Q_1 + Q_2\)
\(Q_{total}=50100J+15675J = 65775J\approx65.8kJ\)

Step1: Calculate heat to raise temperature to \(100^{\circ}C\)

Use formula \(Q = mc\Delta T\). Here \(m = 450g\), \(c = 4.18J/g^{\circ}C\), \(\Delta T=(100 - 20)^{\circ}C\)
\(Q_1=450g\times4.18J/g^{\circ}C\times80^{\circ}C=150480J = 150.48kJ\)

Step2: Calculate heat for vaporization

Let mass of water boiled be \(m\). Heat for vaporization \(Q_2=mH_v\), \(H_v = 2260J/g\)
Total heat \(Q = Q_1+Q_2\). Given \(Q = 325000J\)
\(325000J=150480J+m\times2260J/g\)
\(m\times2260J/g=325000J - 150480J=174520J\)
\(m=\frac{174520J}{2260J/g}\approx77.2g\)

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

Use formula \(Q = mc\Delta T\). Assume \(c = 4.18J/g^{\circ}C\) (similar to water), \(m = 340g\), \(\Delta T=(25 - 0)^{\circ}C\)
\(Q_1=340g\times4.18J/g^{\circ}C\times25^{\circ}C = 35530J\)

Step2: Calculate heat for freezing

Use formula \(Q = mH_f\), \(H_f = 334J/g\)
\(Q_2=340g\times334J/g=113560J\)

Step3: Calculate heat to cool ice to \(- 12^{\circ}C\)

Use formula \(Q = mc\Delta T\). For ice \(c = 2.1J/g^{\circ}C\), \(\Delta T=(0-(-12))^{\circ}C = 12^{\circ}C\)
\(Q_3=340g\times2.1J/g^{\circ}C\times12^{\circ}C=8568J\)

Step4: Calculate total heat

\(Q_{total}=Q_1+Q_2+Q_3\)
\(Q_{total}=35530J + 113560J+8568J=157658J\approx157.7kJ\)

Answer:

\(65.8kJ\)

2.