Let m be the mass of the teapot, c the specific thermal capacity of the substance from which the teapot is made, T the temperature to which the teapot is heated when a mass M of water at temperature T is poured into it, and t the temperature of the room. We shall assume that the thermal capacity of the water remains constant and equal to unity throughout. Let us write down the equation of thermal balance:
mc(T₁-t) = M(T- T₁).
Hence we find that
T₁ = (mct + MT)/(M + mc).
Substituting for all these values from the conditions of the problem, we find: for the copper teapot
T₁ = (200 × 0⋅095 × 20 + 500 × 100)/(500 + 200 × 0⋅095) = 97∘0 approx.
for the china teapot
T₁ = (300 × 0⋅2 × 20 + 500 × 100)/(500 + 300 × 0⋅2) = 91.4°0 approx.
HEAT AND CAPILLARY PHENOMENA 201
Thus, if there were no surface cooling, the copper teapot would be better for making tea. This result is mainly due to the high thermal capacity of china by comparison with that of copper.
If water from the kettle be poured into the teapot before making the tea, the teapot gets hotter and when water is poured into it the second time (for making the tea), it is hotter. As a result of its higher thermal capacity and its lower thermal conductivity the china teapot, when heated first, will grow cold less quickly than the copper one, and therefore in fact it is better.