Question

A 180-g ice cube at 0°C is placed in 900 g of water at 40°C. What is the final temperature of the mixture? (Latent heat=3.33x10⁵ J/Kg, Specific heat=4186 J/Kg °C)

Select one:

a. 12.7°C

b. 10.0°C

c. 20.1°C

d. 17.7°C

e. 15.3°C

Answer 1

The ice qube will melted while the water will cooled. Therefore, we can write the following heat balance equation:

$$m_iL_f+m_ic_i(T_m-t_i)=m_wc_w(t_w-T_m),$$ $$m_iL_f+m_ic_iT_m-m_ic_it_i=m_wc_wt_w-m_wc_wT_m,$$ $$m_ic_iT_m+m_wc_wT_m=m_wc_wt_w+m_ic_it_i-m_iL_f,$$ $$T_m=\dfrac{m_wc_wt_w+m_ic_it_i-m_iL_f}{m_ic_i+m_wc_w},$$

T_m=\dfrac{0.9\ kg\cdot4186\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot40\ ^{\circ}C+0.18\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot0\ ^{\circ}C-0.18\ kg\cdot 3.33\cdot10^5\ \dfrac{J}{kg}}{0.18\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}+0.9\ kg\cdot4186\ \dfrac{J}{kg\cdot \!^{\circ}C}};

$T_m=21.9\ ^{\circ}C,$ none of the above.