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

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:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(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:

$6.07\times c_1\times (47-20.7)=-[26.1\times 1.17\times (47-52.8)]$

$c_1=1.1J/g^oC$

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