Senior Physics - Solutions
## Chapter 10. Heat and Temperature - Worked Solutions

### ...to selected Complex Reasoning questions from the OUP text *"New Century" Senior Physics - Knowledge, Processes and Reasoning* by Walding, Rapkins and Rossiter

**Q 39.**

A rating of 5 kW means it gives heat at the rate of 5 kJ per second

If we consider 1 minute:

5 kW = 5000 J s-1 = 5000 x 60 J min-1 = 300000 J min-1

In 1 minute 5 Kg of water pass through the shower unit.

Q = M c DT.

30000 = 5 x 4180 x DT

DT = 14.4 ^{o}C

The final temperature = 15 ^{o}C + 14.4 ^{o}C = 29.4 ^{o}C

**Q 40**

Assume that the heat lost from the iron goes
into just melting the ice and does not raise its temperature

Q (lost) = Q (gained)

mcDT = mL_{f
}0.1 x 4200 x (150 - 0) = m x 3.34 x 10^{5
}m = 0.1886 kg (= 188.6 g)

Density (D) = m/V

V = m/D = 188.6/7.9 = 23.87 cm^{3
}Length of cylinder = Volume/Area = 23.87/(p
r^{2})

L = 23.87/(p
x 2^{2}) = 1.9 cm

**Q 41**

Assume that the steam cools to 100°C, condenses to water at 100°C and then cools to 0°C. At the same time the ice cube
melts and turns to water at 0°C

Q (lost) = Q (gained)

mcDT + mL_{v} +
mcDT = mL_{f
}m(2.02 x 10 + 2250 + 4.2 x 100) = 20 x 334

m = 2.48 g

\
m (total) = 55 g + 2.48 g = 57.48 g

* Return to Worked Solutions Menu Page page.*

* Return to Objectives-Summary Menu Page page.*

* Return to Physics Textbook Home page*