Formula
The heat needed to change a body's temperature equals specific heat capacity times mass times temperature difference: Q = cmΔT.
ENT-level problem
Water of mass $m_1 = 2$ kg at a temperature of $t_1 = 80°$C is mixed with water of mass $m_2 = 3$ kg at a temperature of $t_2 = 20°$C. Find the temperature of the mixture. Neglect heat loss to the surroundings.
The hot water gives up as much heat as the cold water absorbs (the specific heat capacity is the same and cancels out):
$$c m_1 (t_1 - t) = c m_2 (t - t_2)$$
$$t = \frac{m_1 t_1 + m_2 t_2}{m_1 + m_2} = \frac{2 \cdot 80 + 3 \cdot 20}{2 + 3} = \frac{160 + 60}{5} = 44°\text{C}$$
The heat depends on the substance (specific heat capacity c), the mass of the body and how many degrees its temperature must change. For water c = 4200 J/(kg·K) — which is why it is used in heaters and thermal accumulators.
The temperature difference ΔT is the same in degrees Celsius and kelvins. Heating and cooling use the same formula: for cooling ΔT is taken as an absolute value.
The heat needed to warm 1 kg of a substance by 1 K. For water it is 4200 J/(kg·K), for ice 2100 J/(kg·K).
Find ΔT = Q / (c · m) and add it to the initial temperature: t₂ = t₁ + Q / (c · m).
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