Question

We put 50 g of ice of 0 °C in a beaker containing 450 g of beer of 20 °C, mix, and wait for heat exchange within the beaker. What will be the final temperature?

Answer 1

Answer : The final temperature will be, 292 K

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_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of ice =
`2.05J/g.K`

`c_2` = specific heat of beer =
`4.2J/g.K`

`m_1` = mass of ice = 50 g

`m_2` = mass of beer = 450 g

`T_f` = final temperature = ?

`T_1` = initial temperature of ice =
`0^oC=273+0=273K`

`T_2` = initial temperature of beer =
`20^oC=273+20=293K`

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

`50g* 2.05J/g.K* (T_f-273)K=-450g* 4.2J/g.K* (T_f-293)K`

`T_f=291.971K\approx 292K`

Therefore, the final temperature will be, 292 K