A solar panel transfers energy at a rate of 1.2 kW to liquid passing through it - AQA - A-Level Physics - Question 7 - 2021 - Paper 2

To find the flow rate of the liquid, we can use the formula relating heat transfer, specific heat capacity, mass flow rate, and temperature change:

Q = rac{dm}{dt} imes c imes \Delta T

where:

• $Q$ is the power (in W),
• $\frac{dm}{dt}$ is the mass flow rate (in kg/s),
• $c$ is the specific heat capacity (in kJ/kg·K),
• $\Delta T$ is the change in temperature (in K).

Given that:

• $Q = 1200$ W (1.2 kW),
• $c = 4.0$ kJ/kg·K = 4000 J/kg·K,
• $\Delta T = 3.0$ K,

we can rearrange the equation to solve for the mass flow rate:

$\frac{dm}{dt} = \frac{Q}{c \times \Delta T}$

Substituting in the values:

$\frac{dm}{dt} = \frac{1200 \, \text{W}}{4000 \, \text{J/kg·K} \times 3.0 \, \text{K}}$

Calculating this gives:

$\frac{dm}{dt} = \frac{1200}{12000} = 0.1 \, \text{kg/s}$

Thus, the flow rate of the liquid is 0.10 kg s⁻¹, which corresponds to option A.