A student investigated how much energy from the Sun was incident on the Earth's surface at her location - AQA - GCSE Physics - Question 4 - 2018 - Paper 1

The equation that links energy transferred (E), power (P), and time (t) is given by:

$P = \frac{E}{t}$

To calculate the mean power, we need to use the equation: $P = \frac{E}{t}$

Given that the mass of water is 1050 g (or 1.050 kg) and the specific heat capacity of water is 4200 J/kg°C, we can find the energy transferred:

$E = m \times c \times \Delta T$

Here,

• $m = 1.050 \, kg$
• $c = 4200 \, J/kg°C$
• $\Delta T = 0.6 °C$

Thus,
$E = 1.050 imes 4200 imes 0.6 = 2646 \, J$

Finally, if we assume the time taken is 300 seconds, the mean power is:

$P = \frac{2646}{300} = 8.82 \, W$

Using the equation for energy transfer, we can rearrange it to find the mass:

$m = \frac{E}{c \times \Delta T}$

Plugging in the values:

• $E = 2646 \, J$
• $c = 4200 \, J/kg°C$
• $\Delta T = 0.6 °C$

We have:

$m = \frac{2646}{4200 \times 0.6} = 1.05 \, kg$

Thus, the mass of the water used is approximately 0.417 kg.