Help me please

And can someone give me some tips on how to skor in physics subject

## Question

18

Rajah 13 menunjukkan lengkung pemanasan bagi 0.4 kg cecair yang dipanaskan menggunakan pemanas rendam 100 W.

Diagram 13 shows the heating curve for 0.4 kg liquid heated by 100 W immersion heater.

Berapakah muatan haba tentu cecair itu?

What is the specific heat capacity of the liquid?

A 105 \space Jkg^{-1\space\circ} C^{-1}

B 525 \space Jkg^{-1\space\circ} C^{-1}

C 2625 \space Jkg^{-1\space\circ} C^{-1}

D 4375 \space Jkg^{-1\space\circ} C^{-1}

### Answer

Hi Dhaksheina,

For this question, we are assuming that there is no heat lost to the surroundings as the question did not mention anything about it. So the electrical energy of the heater is converted to heat energy which is used to heat up the water.

We know that Energy=Power \times time, E=Pt, where P refers to the power of the heater which is given in the question, 100 W.

And heat energy, E=mc\theta.

Thus Pt=mc\theta.

From the graph, you can see that after 7 minutes, the temperature of the liquid increases from 30^\circ C to 70^\circ C. So the rise in temperature is 40^\circ C. But for t we have to convert to seconds as it must be in SI unit, so 7 minutes = 7 \times 60 seconds.

Pt=mc\theta

100(7 \times 60)=0.4c(70-30)

42000=16c

c=\frac{42000}{16}=2625Jkg^{-1\space\circ} C^{-1}

Thus the answer should be C.

Hope this helps!