Specific Latent Heat Flashcards
How much energy is required to change a 2.00 kg block of lead to liquid at its melting point?
From Table 2, the specific latent heat of fusion of lead is 2.30 × 104 J kg–1.
Q= m Lf
= 2.00 × 2.30 × 104
= 4.60 × 104 J
An ice tray containing 200 g of water at 25.0°C is placed in the freezer. How much heat
energy has to be removed to change the water into ice at –4.0°C? (Note: cwater = 4180 J kg–1
K–1, cice = 2050 J kg–1 K–1)
You need to recognise that there are three cooling stages in this question:
* heat has to be removed to lower the temperature of water from 25°C to 0°C
* more heat has to be removed to freeze the water at 0°C (its melting point) just to
overcome the kinetic energy keeping the particles apart – this results in ice at 0°C
* even more heat has to be removed to cool the ice down from 0°C to –4°C.
Note that the specific heat capacity of ice (2100 J kg–1 K–1) is different from that of
water (4180 J kg–1 K–1).
Q = mc∆T + m Lf
+ (mc∆T)ice
= 0.200 × 4180 × (25.0 − 0 ) + 0.200 × 3.34 × 105 + 0.200 × 2050 × (4.0 − 0) = 2.1 × 104 + 6.68 × 104 + 1640
= 89.4 × 104 J (3 sf)
A beaker contains 150 mL of water at the boiling point. What extra thermal energy is
required to vaporise it at 100°C?
Q = m Lv
= 0.150 × 2.25 × 106 J kg−1
= 3.38 × 105 J (3 sf)
Determine the energy required to melt 2.50 kg of
gold at its melting point. (Refer to Table 2 for latent
heats of fusion.)
157,500 J
Copper has a melting point of 1083°C. Determine
the energy required to melt 200.0 g of copper
originally at room temperature of 20°C
123,915 J
A 2.0 L bottle of water at 20°C is placed in the
freezer compartment of a refrigerator. Determine
how much thermal energy must be removed by the
refrigerator to freeze this water.
835,200 J
Determine the mass of acetone that can be
vaporised by the addition of 20 kJ of thermal energy
while at its boiling point of 56°C. (acetone Lv
= 5.18 ×
105 J kg–1
38.6 g
