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.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of ice = [tex]2.05J/g.K[/tex]
[tex]c_2[/tex] = specific heat of beer = [tex]4.2J/g.K[/tex]
[tex]m_1[/tex] = mass of ice = 50 g
[tex]m_2[/tex] = mass of beer = 450 g
[tex]T_f[/tex] = final temperature = ?
[tex]T_1[/tex] = initial temperature of ice = [tex]0^oC=273+0=273K[/tex]
[tex]T_2[/tex] = initial temperature of beer = [tex]20^oC=273+20=293K[/tex]
Now put all the given values in the above formula, we get:
[tex]50g\times 2.05J/g.K\times (T_f-273)K=-450g\times 4.2J/g.K\times (T_f-293)K[/tex]
[tex]T_f=291.971K\approx 292K[/tex]
Therefore, the final temperature will be, 292 K
