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 metal plate. The electrical power of the heater inside the iron is 2.1 kW. Assume that all the energy from the heater is transferred to the metal plate. The metal plate has a mass of 1.2 kg and is initially at a temperature of 20 °C. The heater is switched on. After a time t, the metal plate reaches its working temperature of 125 °C. Calculate t. The metal plate is maintained at its working temperature. The power at 20 °C drips continuously onto the metal plate. Steam at 100 °C emerges continuously from the iron. The maker claims that the iron can generate steam at a rate of 60 g min⁻¹. Determine whether this claim is true. The specific latent heat of vaporisation of water = 2.3 × 10⁵ J kg⁻¹ The specific heat capacity of water = 4200 J kg⁻¹ K⁻¹

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

Question 1

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:

$Q = mc\Delta T$

Where:

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

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

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

$t = \frac{Q}{P} = \frac{63000}{2100} = 30 \, \text{s}$

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 = $\frac{0.06}{60} = 0.001 \text{kg s}^{-1}$ .

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

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

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

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

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

Calculate t.

Determine whether this claim is true.

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