1. The heat energy required to melt 0.3 kg of ice at 0°C is calculated using the formula Q = m × L_f, where m is mass and L_f is the specific latent heat of fusion.
Q = 0.3 kg × 3.34 × 10⁵ J/kg = 1.002 × 10⁵ J.
2. The specific latent heat of fusion (L_f) is found by dividing the total heat given out by the mass: L_f = Q / m.
L_f = (6 × 10⁵ J) / 4.0 kg = 1.5 × 10⁵ J/kg.
3. To find the mass of water vaporized, use Q = m × L_v, rearranged to m = Q / L_v.
m = (10.6 × 10³ J) / (2.26 × 10⁶ J/kg) = 0.00469 kg or 4.69 g.
4a. The heat required to vaporize 1 g (0.001 kg) of ammonia is Q = m × L_v.
Q = 0.001 kg × 1.34 × 10⁶ J/kg = 1340 J.
4b. This heat (1340 J) is used to freeze water at 0°C. Using Q = m × L_f, solve for m: m = Q / L_f.
m = 1340 J / (3.34 × 10⁵ J/kg) = 0.00401 kg or 4.01 g of water frozen per gram of ammonia vaporised.
5a. Region AB: Solid wax is heating up. Region BC: Wax is melting (phase change, temperature constant). Region CD: Liquid wax is heating up.
5b. In region BC, the wax melts. Time duration is from 50 s to 100 s, so Δt = 50 s. Power P = 200 W, so energy supplied Q = P × Δt = 200 W × 50 s = 10,000 J. For a 2 kg lump, L_f = Q / m = 10,000 J / 2 kg = 5000 J/kg.
6. Using conservation of energy, heat lost by warm water = heat gained by cold water.
Let final temperature be T.
(0.2 kg)(4180 J/kg°C)(40 - T) = (0.1 kg)(4180 J/kg°C)(T - 10)
Divide both sides by 4180: 0.2(40 - T) = 0.1(T - 10)
8 - 0.2T = 0.1T - 1
9 = 0.3T
T = 30°C.
Parent Tip: Review the logic above to help your child master the concept of latent heat worksheet.
