Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 13.7 °C are mixed. The specific heat of water is 4.184 J/g°C.

Answer 1

The final temperature of the system is 42.46°C.

Step-by-step explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c* (T_f-T_1)=-(m_2* c* (T_f-T_2))`

where,

c = specific heat of water=
`4.18J/g^oC`

`m_1` = mass of water sample with 100 °C= 50.0 g

`m_2` = mass of water sample with 13.7 °C= 100.0 g

`T_f` = final temperature of system

`T_1` = initial temperature of 50 g of water sample=
`100^oC`

`T_2` = initial temperature of 100 g of water =
`13.7^oC`

Now put all the given values in the given formula, we get

`50.0 g* 4.184 J/g^oC* (T_f-100^oC)=-(100 g* 4.184 J/g^oC* (T_f-13.7^oC))`

`T_f=42.46^oC`

The final temperature of the system is 42.46°C.