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Figure 1 shows an electric steam iron - AQA - A-Level Physics - Question 1 - 2022 - Paper 2

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Figure 1 shows an electric steam iron. Water from a reservoir drips onto an electrically-heated metal plate. The water boils and steam escapes through holes in the ... show full transcript

Worked Solution & Example Answer:Figure 1 shows an electric steam iron - AQA - A-Level Physics - Question 1 - 2022 - Paper 2

Step 1

Calculate t.

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Answer

To find the time t taken for the metal plate to reach the desired temperature, we use the equation for heat transfer, where:

Q=mcΔTQ = mc \Delta T

Here,

  • QQ is the heat energy supplied,
  • m=1.2 kgm = 1.2 \, kg (mass of the metal plate),
  • c=450 J kg−1 K−1c = 450 \, J \, kg^{-1} \, K^{-1} (specific heat capacity),
  • DeltaT=Tfinal−Tinitial=125∘C−20∘C=105 K\\Delta T = T_{final} - T_{initial} = 125^{\circ}C - 20^{\circ}C = 105 \, K.

Calculating the heat energy:

Q=1.2 kg×450Jkg K×105 K=63000 JQ = 1.2 \, kg \times 450 \frac{J}{kg \, K} \times 105 \, K = 63000 \, J

The power supplied by the heater is given as 2.1 kW:

Power=2.1 kW=2100 WPower = 2.1 \, kW = 2100 \, W

The time taken to transfer this energy can be found using:

t=QP=63000 J2100 W=30 st = \frac{Q}{P} = \frac{63000 \, J}{2100 \, W} = 30 \, s

Step 2

Determine whether this claim is true.

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Answer

To evaluate the claim, we will find out if the iron can generate steam at a rate of 60 g min–1.

First, convert the mass flow rate to kg/s:

60g/min×1 kg1000 g×1 min60 s=0.001 kg/s60 g/min \times \frac{1 \, kg}{1000 \, g} \times \frac{1 \, min}{60 \, s} = 0.001 \, kg/s

The energy required to convert the mass of water to steam can be calculated using:

E=mLE = mL

Where:

  • m=0.001 kg/sm = 0.001 \, kg/s (mass flow rate),
  • L=2.3×106 J/kgL = 2.3 \times 10^6 \, J/kg (specific latent heat of vapourisation).

Thus:

E=0.001 kg/s×2.3×106 J/kg=2300 J/s=2300 WE = 0.001 \, kg/s \times 2.3 \times 10^6 \, J/kg = 2300 \, J/s = 2300 \, W

Since the power provided by the heater is only 2100 W, we see that the power required (2300 W) is greater than the output power of the iron.

Therefore, the maker's claim that the iron can generate steam at a rate of 60 g min–1 is false.

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