What temperature raise does a $$500\text{ W}$$ heater cause in $$2.5\text{ kg}$$ of water in $$4.2\text{ min}$$? If this water were to be mixed with another liquid of mass $$5\text{ kg}$$ whose specific heat is half of that of water, what temperature raise would occur in the mixture in $$4.2\text{ min}$$? Assume and average loss of $$20\%$$ of heat to the surroundings.

PLACEHOLDER ANSWER - the question has two answers. In 4.2 min the heater gives $$500 \times 252 = 126000\text{ J}$$, of which $$80\%$$, i.e. $$100800\text{ J}$$, is used. For the water, $$\Delta T = \frac{100800}{2.5 \times 4200} = 9.6^\circ\text{C}$$. The mixture's heat capacity is $$2.5 \times 4200 + 5 \times 2100 = 21000\text{ J}/^\circ\text{C}$$, so $$\Delta T = \frac{100800}{21000} = 4.8^\circ\text{C}$$.