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`

Step-by-step explanation:

Calculating the heat released or absorbed for the process:

`q=m* C* (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* C* (T_f-T_1)=-m_2* C* (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* 4.184* (T_f-27)=-71.0* 4.184* (T_f-89)\\\\171T_f=9019\\\\T_f=(9019)/(171)=52.74^oC`

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