Elon took 300g of water and supplied 8000 joules of heat to it.

Question

What is the change in temperature of the water? (Specific heat capacity of water is 4.2 /g°C^-1)

Answer 1

We can simply use the following equation:

q = mCΔT

From here, we just rearrange and solve for the change in the temperature:

q = mCΔT => ΔT = q/mC

<=> 8000J/(300g*4.2 J/g°C)

ΔT = 6.349 °C

Cheers!