Question

ImTwo substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.44 g and its initial temperature is 20.3 oC. The mass of substance B is 25.1 g and its initial temperature is 52.8 oC. The final temperature of both substances at thermal equilibrium is 46.7 oC. If the specific heat capacity of substance B is 1.17 J/g⋅oC, what is the specific heat capacity of substance A

Answer 1

Answer:

Specific heat capacity of A is 1.05J/g°C

Explanation:

Hello,

In this question, we have to bodies in contact and there's transfer of energy (heat) between them. The heat loss in one body is equal to the heat gain in another body.

Heat loss in B = Heat gain in A

Data,

Mass of A = 6.44g

Initial temperature of A (T1) = 20.3°C

Mass of B = 25.1g

Initial temperature of B = 52.8°C

Final temperature after equilibrium is achieved (T3) = 46.7°C

Specific heat capacity of A (C1) = 1.17J/g°C

Specific heat capacity of B (C2) = ?

Heat loss in B = Heat gain in A

Q = mc∇T

Q = heat energy

M = mass of substance

C = specific heat capacity of the substance

∇T = change in temperature of the substance

M₂C₂(T₂ - T₃) = M₁C₁(T₃ - T₁)

Substitute the values into the variables and solve for C₁

25.1 × 1.17 × (52.8 - 46.7) = 6.44 × C₁ × (46.7 - 20.3)

179.1387 = 170.016C₁

Divide both sides by 170.016

C₁ = 179.1387 / 170.016

C₁ = 1.05J/g°C

The specific heat capacity of A is 1.05J/g°C