Answer to Question #162097 in Physics for Blessed Joy

Question

 A thirsty farmer cools a 1.00-L bottle of softdrink (mostly water) by pouring the

contents into a large copper mug with a mass of 0.278 kg and adding 0.0580 kg of ice initially at -16^oC. If the softdrink and mug are initially at 20^oC, what is the final temperature of the system, assuming no heat losses?

Answer 1

Let us find the final temperature t.

Specific heat for ice is 2100 J/kg/°C, for water 4200, for copper 385. Latent heat for ice is 334000 J/kg.

Heat lost by softdrink and mug:

$$Q_1=c_sm_s(t-t_s)+c_cm_c(t-t_s),$$

heat received by ice:

$$Q_2=c_im_i(0-t_i),$$

heat of fusion for ice:

$$Q_3=m_i\lambda,$$

heat to warm the water (when ice melt) from 0°C to $t$:

$$Q_4=c_sm_i(t-0).$$

Thermal equilibrium:

$$Q_1+Q_2+Q_3+Q_4=0,\\ c_sm_s(t-t_s)+c_cm_c(t-t_s)+m_i\lambda+c_sm_it-c_im_it_i=0,\\\space\\ t=\frac{c_sm_st_s+c_cm_ct_s +c_im_it_i-m_i\lambda}{c_sm_s+c_cm_c+c_sm_i}=\\\space\\ =14º\text{C}.$$