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
