Two liquids of the same mass (m1 = m2), but having different specific heat capacities (с1 = 2с2) and

Two liquids of the same mass (m1 = m2), but having different specific heat capacities (с1 = 2с2) and different initial temperatures (T1 = 2Т2) were mixed in the calorimeter. Determine the steady-state temperature of the mixture (neglecting heat loss).

Given:

m1 = m2

c1 = 2 * c2

T1 = 2 * T2

To find:

T -?

Decision:

According to the heat balance equation, Delta Q1 + Delta Q2 = 0,

where Delta Q1 = c1 * m1 * Delta T1 is the amount of heat received by a liquid with a mass of m1;

Delta Q2 = c2 * m2 * Delta T2 – the amount of heat given off by a liquid with a mass of m2.

Let the steady-state temperature of the mixture be equal to T. Then Delta T1 = T – T1,

and Delta T2 = T2 – T. Therefore, c1 * m1 * (T – T1) = c2 * m2 * (T2 – T). From where

T = (c1 * m1 * T1 + c2 * m2 * T2) / (c1 * m1 + c2 * m2).

Substituting the known data, we find T:

T = (2 * c2 * m2 * 2 * T2 + c2 * m2 * T2) / (2 * c2 * m2 + c2 * m2) = (5 * c2 * m2 * T2) / (3 * c2 * m2) = 5 * T2 / 3.

Answer: T = 5 * T2 / 3.