Question

A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the water? (Specific heat capacity of water = 4.184 J/goC.)

Answer 1

Answer: The final temperature will be $52.74^oC$

Explanation:

Calculating the heat released or absorbed for the process:

$q=m\times C\times (T_2-T_1)$

In a system, the total amount of heat released is equal to the total amount of heat absorbed.

$q_1=-q_2$

OR

$m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2)$ ......(1)

where,

C = heat capacity of water = $4.184J/g^oC$

$m_1$ = mass of water of sample 1 = 100.0 g

$m_2$ = mass of water of sample 2 = 71.0 g

$T_f$ = final temperature of the system = ?

$T_1$ = initial temperature of water of sample 1 = $27^oC$

$T_2$ = initial temperature of the water of sample 2 = $89.0^oC$

Putting values in equation 1, we get:

$100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC$

Hence, the final temperature will be $52.74^oC$