1. The problem statement, all variables and given/known data 380g of a liquid at 12'C in a copper calorimeter weighing 90g is heating at a rate of 20 watt for exactly 3 minutes to produce a temperature of 17'C. If the specific heat capacity of copper is 40Jkg-1K-1, the thermal capacity of the heater is negligible, and there is a negligible heat loss to the surroudnings, obtain a value for the specific heat capacity of the liquid. 2. Relevant equations VxI = MC(x2 - x1) + q 3. The attempt at a solution I genuienly don't have a clue about this. +q is negligible. So what are we meant to do for this? I assume we get two values for the equation of the calorimeter and the liquid, minus them and make it C =. Is that correct? But what values do we sub in for where (I know what everything stands for =/). Any help is apprectiated! Thank you.