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.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

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:

$50g\times 2.05J/g.K\times (T_f-273)K=-450g\times 4.2J/g.K\times (T_f-293)K$

$T_f=291.971K\approx 292K$

Therefore, the final temperature will be, 292 K