5/2/2016 11:00 PM A s4.2/100 5/2/2016 05:55 PM Gradebook Print Calculator Period

Question

5/2/2016 11:00 PM A s4.2/100 5/2/2016 05:55 PM Gradebook Print Calculator Periodic Table Question 13 of 16 Map A Presented by Physics openstax sapling learning How much total heat transfer is necessary to lower the temperature of 0.335 kg of steam from 129.5 °C to 19.5 °C, including the energy for phase changes? Number How much time is required for each stage of this process, assuming a constant 835.0 Wrate of heat exchange? Give the times in the order that the stages occur Number Number Number Number Number Hint O Previous ® Give Up & View Solution Check Answer Next Ext

Explanation / Answer

m = 0.335 Kg = 335 g
Ti = 129.5 oC
Tf = -19.5 oC

Specific Heat of Water, Cw = 4.186 J/gm k
Specific Heat of ice, Ci = 2.108 J/gm k
Specific Heat of Steam, Cs = 1.996 J/gm k
Latent Heat of ice, Li = 333.55 J/gm
Latent Heat of Vaporisation, Ls = 2257 J/gm

Stage 1, from 129.5 oC  to 100 oC
Q1 = 335 * 1.996 * (129.5 - 100) J
Q1 = 19725.5 J

Energy = Power * time
19725.5 = 835.0 * t1
t1 = 23.6 s

Stage 2, from 100 oC steam to water.
Q2 = 335 * 2257 J
Q2 = 756095 J

t2 = 756095/835
t2 = 905.5 s

Stage 3, from 100 oC to 0 oC .
Q3 = 335 * 4.186 * 100 J
Q3 = 140231 J

t3 = 140231/835
t3 = 167.9 s

Stage 4, from 0 oC water to ice.
Q4 = 335 * 334 J
Q4 = 111890 J

t4 = 111890/835
t4 = 134 s

Stage 5, from 0 oC to - 19.5 oC
Q5 = 335 * 2.108 * 19.5 J
Q5 = 13770.5 J

t5 = 13770.5/835
t5 = 16.49 s

Total Heat Transfer, Q = Q1 + Q2 + Q3 + Q4 + Q5
Total Heat Transfer, Q = 19725.5 + 756095 + 140231 + 111890 + 13770.5
Total Heat Transfer, Q = 1041712 J

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.