The calorimeter contains V1 = 1 liter of water at a temperature t1 = 14∘C. It is filled with V2 = 5 liters of water

The calorimeter contains V1 = 1 liter of water at a temperature t1 = 14∘C. It is filled with V2 = 5 liters of water with a temperature of t2 = 50∘C. What temperature is set in the calorimeter? Express the answer in ∘C, rounded to the nearest whole number. Disregard the heat capacity of the calorimeter. Neglect the loss of heat.

V1 = 1 l = 1 * 10 ^ -3 m ^ 3.

t1 = 14 ∘C.

V2 = 5 L = 5 * 10 ^ -3 m ^ 3.

t2 = 50 ∘C.

C = 4200 J / kg * ∘C.

ρ = 1000 kg / m ^ 3.

t -?

When mixed, water with a higher temperature will give up a certain amount of heat Q, which will be used to heat water with a lower temperature.

Q = C * m2 * (t2 – t).

Q = C * m1 * (t – t1).

Let’s write down the heat balance equation: С * m2 * (t2 – t) = С * m1 * (t – t1).

The mass of water is determined by the formula: m1 = ρ * V1, m2 = ρ * V2.

С * ρ * V2 * (t2 – t) = С * ρ * V1 * (t – t1);

V2 * (t2 – t) = V1 * (t – t1);

V2 * t2 – V2 * t = V1 * t – V1 * t1;

V2 * t2 + V1 * t1 = V2 * t + V1 * t;

t = (V2 * t2 + V1 * t1) / (V2 + V1).

t = (5 * 10 ^ -3 m ^ 3 * 50 ∘C. + 1 * 10 ^ -3 m ^ 3 * 14 ∘C) / (5 * 10 ^ -3 m ^ 3 + 1 * 10 ^ -3 m ^ 3) = 44 ∘C.

Answer: the calorimeter will set the temperature t = 44 ∘C.