2018 A/L Physics Past Paper · Question 20 (English Medium)

Power of the immersion heater

Water at 0 ∘C0\ ^\circ\text{C} is continuously fed into a fully insulated boiler fixed with an immersion heater at a constant rate of 1×10−21 \times 10^{-2} kg s⁻¹. The specific heat capacity and the specific latent heat of vaporization of water are 4.2×1034.2 \times 10^3 J kg⁻¹ °C⁻¹ and 2.25×1062.25 \times 10^6 J kg⁻¹ respectively. If the steam at 100 ∘C100\ ^\circ\text{C} is to be produced at the same rate as that of supply of water, the power of the immersion heater should be

  1. (1) 4.2 kW
  2. (2) 22.5 kW
  3. (3) 26.7 kW
  4. (4) 42.0 kW
  5. (5) 267.0 kW
Answer and explanation

Answer: (3) 26.7 kW

Given information

We know: water comes in at m˙=1×10−2 kg s−1\dot m = 1 \times 10^{-2}\ \text{kg s}^{-1}.

It comes in at 0 ∘C0\ ^\circ\text{C} and leaves as steam at 100 ∘C100\ ^\circ\text{C}.

c=4.2×103 J kg−1 ∘C−1c = 4.2 \times 10^{3}\ \text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1} and L=2.25×106 J kg−1L = 2.25 \times 10^{6}\ \text{J kg}^{-1}.

The boiler is fully insulated. No heat is lost.

What the question asks

We want: the power PP of the immersion heater.

Relevant theory

The amount of heat required to raise the temperature of a unit mass of a substance by 1 unit of temperature (1 °C or 1 K) defined as the specific heat capacity of that substance.

Specific latent heat of vapourization of a liquid is defined as the heat required to change the state of a unit mass of a liquid to its vapour without change of temperature (at its boiling point).

Required equations

P1=m˙cΔTP_1 = \dot{m} c \Delta T

P2=m˙LP_2 = \dot{m} L

P=P1+P2P = P_1 + P_2

Applying it to this question

There are two parts.

First, warm the water from 0 ∘C0\ ^\circ\text{C} to 100 ∘C100\ ^\circ\text{C}.

Then turn it into steam at 100 ∘C100\ ^\circ\text{C}.

We work per second. So each energy becomes a power.

Reasoning and calculation

4.2×1034.2\times10^3

P1=10−2×4.2×103×100P_1 = 10^{-2}\times4.2\times10^3\times100

P1=4.2 kWP_{1} = 4.2\ \text{kW} only warms the water.

The same water must still turn into steam.

2.25×1042.25\times10^4

P2=10−2×2.25×106P_2 = 10^{-2}\times2.25\times10^6

2.67×1042.67\times10^4

P=4.2×103+2.25×104P = 4.2\times10^3 + 2.25\times10^4

Checks

Check the unit.

m˙cΔθ\dot m c\Delta\theta: kg s−1×J kg−1 ∘C−1×∘C\text{kg s}^{-1} \times \text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1} \times {}^{\circ}\text{C} gives J s−1\text{J s}^{-1}, that is W.

m˙L\dot m L: kg s−1×J kg−1\text{kg s}^{-1} \times \text{J kg}^{-1} gives W.

Both are in W. So we can add them.

mdot∗(c∗dT+L)mdot * (c * dT + L)

Does it make sense?

Boiling takes 22.5 kW. Warming takes only 4.2 kW.

For water, LL is much bigger than c×100c \times 100. So boiling needs about five times more.

Conclusion

P=26.7 kWP = 26.7\ \text{kW}

Warming: 4.2 kW.

Boiling: 22.5 kW.

P=4.2+22.5=26.7 kWP = 4.2 + 22.5 = 26.7\ \text{kW}

The answer is option (3).

P = ṁL only

The water comes in at 0 ∘C0\ ^\circ\text{C}.

So it must first be warmed to 100 ∘C100\ ^\circ\text{C}.

P=m˙(cΔθ+L)P = \dot m(c\Delta\theta + L)

Explore

Water at 0 ∘C0\ ^\circ\text{C} runs into the boiler, 0.01 kg every second.

The heater first warms it to 100 ∘C100\ ^\circ\text{C}. That needs 4.2 kW.

Then it turns the water into steam. That needs 22.5 kW.

Both happen all the time. So the heater needs 4.2+22.5=26.74.2 + 22.5 = 26.7 kW.

தமிழ் · සිංහල