Question

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.07 g and its initial temperature is 20.7 ∘C. The mass of substance B is 26.1 g and its initial temperature is 52.8 ∘C. The final temperature of both substances at thermal equilibrium is 47.0 ∘C. Part A If the specific heat capacity of substance B is 1.17 J/g⋅∘C, what is the specific heat capacity of substance A? Express your answer using two significant figures.

Answer 1

Try this suggested solution (the figures are not provided).

The answer is marked with red colour.

Answer 2

Answer: The specific heat of substance A is 1.1 J/g°C

Step-by-step explanation:

When substance A is mixed with substance B, the amount of heat released by substance B (initially present at high temperature) will be equal to the amount of heat absorbed by substance A (initially present at low temperature)

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of substance A = 6.07 g

`m_2` = mass of substance B = 26.1 g

tex]T_{final}[/tex] = final temperature = 47.0°C

`T_1` = initial temperature of substance A = 20.7°C

`T_2` = initial temperature of substance B = 52.8°C

`c_1` = specific heat of substance A = ?

`c_2` = specific heat of substance B = 1.17 J/g°C

Putting values in equation 1, we get:

`6.07* c_1* (47-20.7)=-[26.1* 1.17* (47-52.8)]`

`c_1=1.1J/g^oC`

Hence, the specific heat of substance A is 1.1 J/g°C