Question

How much total heat transfer is necessary to lower the temperature of 0.295 kg of steam from 133.5 °C to -31.5 °C, including the energy for phase changes? total heat transfer: How much time is required for each stage of this process, assuming a constant 815.0 W rate of heat exchange? Give the times in the order that the stages occur. 1 / Question Credit: OpenStax College Physics

Answer 1

Let , c_s = 1996 J/(kg.K) = specific heat capacity of steam

c_w = specific heat capacity of water = 4186 J/(kg.K)

c_i = specific heat capcity of ice = 2100 J/(kg.K)

L_s = 2.256×10⁶ J/kg = latent heat of vapourisation of steam

L_f = 3.35×10⁵ J/kg = latent heat of fusion of water

Q₁ = Amount of heat extracted as the temperature drops from 133.5°C to 100°C

Q₁ = mc_s$\Delta$T , where $\Delta$ T is difference in temperature

Q₁ = 0.295×1996×33.5

Q₁ = 19725.47 J

Time taken in the process t₁ = 19725.47/815

t₁ = 24.20 s

Q₂ = amount of heat extracted as phase changes from steam to water

Q₂ = mL_s

Q₂ = 0.295× 2.256×10⁶

Q₂ = 6.65 × 10⁵ J

Time taken in the process

t₂ = 6.65 × 10⁵ / 815

t₂ = 815.95 s

Q₃ = amount of heat extracted as temperature of water drops from 100°C to 0°C

Q₃ = mc_w$\Delta$T

Q₃ = 0.295 × 4186×100

Q₃ = 123487 J

Time taken in this process

t₃ = 123487/815

t₃ = 151.52 s

Q₄ = amount of heat extracted as phase changes from water to ice = mL_s

Q₄ = 0.295 x 3.35 × 10⁵ = 98825 J

Time taken t₄ in this process = 98825 / 815

t₄ = 121.26 s

Q₅ = amount of heat extracted as temperature of ice drops from 0°C to - 31.5°C

Q_{5 =} mc_i$\Delta$T

Q₅ = 0.295×2100×31.5

Q₅ = 19514.25 J

Time taken in the process

t₅ = 19514.25 / 815

t₅ = 23.93

Let, total heat energy extracted is given by

Q = Q₁ + Q₂ + Q₃ + Q₄ 7

Q = 19725.47 + 665000 + 123487 +98825 + 19514.25

Q = 926551.72

Q = 9.26 × 10⁵ J

Various times are

t₁ = 24.20 s

t₂ = 815.95 s

t₃ = 151.52 s

t₄ = 121.26 s

t₅ = 23.93 s