Question

1. 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?
2. 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 energy required to heat water from \(20^{\circ}C\) to \(100^{\circ}C\)

The specific heat capacity of water \(c = 4.18\space J/g^{\circ}C\), mass of water \(m_1=450\space g\), initial temperature \(T_1 = 20^{\circ}C\), final temperature \(T_2=100^{\circ}C\).
Using the formula \(Q_1=mc\Delta T\), where \(\Delta T=T_2 - T_1\)
\(Q_1=450\times4.18\times(100 - 20)\)
\(Q_1=450\times4.18\times80\)
\(Q_1 = 150480\space J=150.48\space kJ\)

Step2: Calculate the energy available for vaporization

Total energy \(Q_{total}=325\space kJ\)
Energy for vaporization \(Q_2=Q_{total}-Q_1\)
\(Q_2=325 - 150.48=174.52\space kJ = 174520\space J\)

Step3: Calculate the mass of water vaporized

The latent heat of vaporization of water \(H_v = 2260\space J/g\)
Using the formula \(Q_2 = mH_v\), then \(m=\frac{Q_2}{H_v}\)
\(m=\frac{174520}{2260}\approx77.2\space g\)

Answer:

The mass of water boiled away is approximately \(77.2\space g\)