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) hv
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:

Step1: Calculate the heat required to melt the ice

The formula for heat of fusion is \(Q = mH_f\). Here, \(m = 150\ g\) and \(H_f=334\ J/g\).

$$Q_1=150\times334 = 50100\ J$$

Step2: Calculate the heat required to warm the water

The formula for heat transfer is \(Q = mc\Delta T\). Here, \(m = 150\ g\), \(c = 4.18\ J/g^{\circ}C\), and \(\Delta T=25 - 0=25^{\circ}C\)

$$Q_2=150\times4.18\times25=15675\ J$$

Step3: Calculate the total heat

$$Q_{total}=Q_1 + Q_2$$

$$Q_{total}=50100+15675 = 65775\ J$$

Convert to \(kJ\): \(Q_{total}=\frac{65775}{1000}=65.8\ kJ\)

Answer:

\(65.8\ kJ\)