A level thermal physics

0.38 kg of a liquid at 12 °C is heated in a copper can weighing 0.9 kg by an electrical heater of power 20 W for 3 minutes. It reaches a temperature of 17 °C. Assuming no heat losses to the surroundings, calculate a value for the specific heat capacity of the liquid.

Reply 1

Applesauwce

initially you'd solve for c and since E=Pt and we take the liquids details:
mcT=Pt
0.38 x c x (17-12)= 20 x (3x60)
1.9c=3600
c=1894.7 (1dp)

but since they've mentioned the can to be made of copper you'll have to take that into consideration, we know exactly how much energy is being used to heat the copper if you look up the SHC of Cu you'll get 385 J/K/kg, assuming that the change in temperature is 5K and the mass is 0.9kg, you can now calculate how much energy has been used to heat the copper, and therefore how much energy has been used to heat the liquid, and from that solve for the liquid's SHC.