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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 calculate the time taken to heat the metal plate to its working temperature, we use the equation:

Q=mcΔTQ = mc\Delta T

Where:

  • m=1.2 kgm = 1.2 \, \text{kg} (mass of the metal plate)
  • c=450 J kg−1K−1c = 450 \, \text{J kg}^{-1} \text{K}^{-1} (specific heat capacity of the metal)
  •  ΔT=125−20=105 K\, \Delta T = 125 - 20 = 105 \, \text{K} (change in temperature)

Calculating the heat energy required: Q=1.2imes450imes105=63000 JQ = 1.2 imes 450 imes 105 = 63000 \, \text{J}

The electrical power supplied is 2.1 kW, which is equal to 2100 W. The time can be calculated using:

t=QP=630002100=30 st = \frac{Q}{P} = \frac{63000}{2100} = 30 \, \text{s}

Step 2

Determine whether this claim is true.

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Answer

To determine the validity of the claim, we first convert the steam generation rate from grams to kilograms:

Rate of steam generation = 60 g min⁻¹ = 0.06 kg min⁻¹ = 0.0660=0.001kg s−1\frac{0.06}{60} = 0.001 \text{kg s}^{-1}.

The energy needed to generate this steam is: Qsteam=m×L=0.001×2.3×105=230 J s−1Q_{steam} = m \times L = 0.001 \times 2.3 \times 10^5 = 230 \, \text{J s}^{-1}

The energy that can be supplied by the heater: P=2100 WP = 2100 \, \text{W}

Since PP is greater than QsteamQ_{steam}, the claim is valid. Therefore, the iron can generate steam at the claimed rate of 60 g min⁻¹.

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