Question

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.01 g and its initial temperature is 20.0 degrees Celsius. The mass of substance B is 25.6 g and its initial temperature is 52.2 degrees Celsius. The final temperature of both substances at thermal equilibrium is 46.1 degrees Celsius. If the specific heat capacity of substance B is 1.17 J/g degrees Celsius, what is the specific heat capacity of substance A

Answer 1

The specific heat capacity of substance A is 1.16 J/g

Step-by-step explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

FOR SUBSTANCE A:

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

FOR SUBSTANCE B:

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

C = 1.16 J/g