Answer to Question #162100 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

Given

Specific heat of water: 4182 J/kg/°C

Specific heat of ice: 2040 J·kg/°C

Specific heat of copper: 385 J/kg/°C

Latent heat of fusion for ice: 334000 J/kg

Temperature of softdrink: $t_s=20º\text{C}$

Temperature of copped mug: $t_s=20º\text{C}$

Temperature of ice: $t_s=-16º\text{C}$

Find

Final temperature of thermal equilibrium $t$

Solution

Put that

$Q_1=c_sm_s(t-t_s)+c_cm_c(t-t_s)$ is the amount of heat lost by the softdrink and mug;

$Q_2=c_im_i(0-t_i)$ is the amount of heat received by ice while it warmed from -16º to 0ºC;

$Q_3=m_i\lambda$ is the amount of heat spent to melt the ice;

$Q_4=c_sm_i(t-0)$ is the amount of heat to warm the melted ice from 0º to the final temperature.

Thermal equilibrium:

$$Q_1+Q_2+Q_3+Q_4=0,\\ c_sm_s(t-t_s)+c_cm_c(t-t_s)+m_i\lambda+c_sm_it-c_im_it_i=0,\\\space\\ t=\frac{c_sm_st_s+c_cm_ct_s +c_im_it_i-m_i\lambda}{c_sm_s+c_cm_c+c_sm_i}=\\\space\\ =14.2º\text{C}.$$