Answer to Question #273487 in Molecular Physics | Thermodynamics for renald

Question

Q 5. A 5 kg of ice cube at −10℃ is mixed with 0.1 kg of water at 80 ℃. There is no heat 

loss to the surrounding. The specific heat capacity of ice is 2.22 𝑘𝐽𝐾

−1𝑘𝑔−1

,The specific 

heat capacity of water is 4.187 𝑘𝐽𝐾

−1𝑘𝑔−1

. The specific latent heat of fusion of ice is 

333 𝑘𝐽 𝑘𝑔−1

.

(a) What is the final physical state of the mixture? (Gas, Liquid or Solid?)

Justify your answer using suitable calculations 

(b) What is the final temperature of the mixture?

Answer 1

Solution;

Given;

For water;

$m_w=0.1kg$

$C_{p_w}=4.187kJ/kgK$

$T_w=80°c=353K$

For ice;

$m_i=5kg$

$T_i=-10°c=263K$

$C_{p_i}=2.22kJ/kgK$

$L_f=333kJ/kg$

(a)

Final physical state is solid.

To justify we find the final temperature;

(b)

Heat loss by the water is heat gained by the ice;

Heat lost by water;

$Q_w=m_wC_wT_f-m_wC_wT_w$

$Q_w=(0.1×4.187)T_f-(0.1×4.187×353)$

$Q_w=0.4187T_f-147.8011$

Heat gained by ice;

$Q_i=m_iC_{p_w}T_f-m_iC_{p_i}T_i-m_iL_f$

$Q_i=(5×4.187)T_f-(5×2.22×263)-(5×333)$

$Q_i=20.935T_f-4584.3$

Therefore;

If;

$Q_i=Q_w$

$20.935T_f-4584.3=0.4187T_f-147.8011$

$20.5163T_f=4436.4989$

$T_f=216.27K$

Therefore the state of mixture is solid since the final temperature is lower than the melting point of ice.