Temperature change of water and the copper can

The question is:

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

Shc of can is 400Jkg^-1K^-1

So far I have the following equation:
DeltaQcan + DeltaQliquid = 20360

Using Q=mcDeltaT,
DeltaQcan = mcanccanDeltaTcan = 0.9400DeltaTcan

DeltaQliquid = 0.38cliquid5

The answer says cliquid = 946.4Jkg^-1K^-1, which you can achieve through saying DeltaTcan = 5, hence:
0.94005 + 0.38cliquid5 = 20360, solving for 947.4.

However, I am unsure about the temperature change of the can. It doesn’t say that the liquid is in thermal equilibrium with the can after the three minutes, so why can we assume that the can has the same temperature change as the liquid? Why can the can not be hotter than the water after those three minutes?