To what temperature will 10 liters of boiling water, taken at a temperature of 100 degrees, cool down
To what temperature will 10 liters of boiling water, taken at a temperature of 100 degrees, cool down, releasing 1680 kJ of energy into the surrounding space?
V = 10 l = 10 * 10 ^ -3 m ^ 3.
C = 4200 J / kg * ° C.
t = 100 ° C.
ρ = 1000 kg / m ^ 3.
Q = 1680 kJ = 1680 * 10 ^ 3 J.
The amount of heat Q that is released when the substance cools is determined by the formula: Q = C * m * (t – t1), where C is the specific heat capacity of the substance, m is the mass of the substance, t, t1 are the initial and final temperatures of the substance.
We express the mass of a substance m by the formula: m = ρ * V, where ρ is the density of the substance, V is the volume of the substance.
Q = C * ρ * V * (t – t1).
t1 = t – Q / C * ρ * V.
t1 = 100 ° C – 1680 * 10 ^ 3 J / 4200 J / kg * ° C * 1000 kg / m ^ 3 * 10 * 10 ^ -3 m ^ 3 = 60 ° C.
Answer: the water temperature will become t1 = 60 ° C.