Question

Four ice cubes at exactly 0 ∘c with a total mass of 52.5 g are combined with 160 g of water at 90 ∘c in an insulated container. (δh∘fus=6.02 kj/mol, cwater=4.18j/g⋅∘c). if no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

Heat gained by ice cubes would be equal to the - heat lost by warm water 

The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol 

Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J 

Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C 

In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature: 

q(ice/water) = - q(warm water) 

moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1) 

50.5 g / 18.0 g/mol = 2.81 mol 

2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80) 

16916 + 211.3T2 = -669.4 T2 + 53555 

36639 = 880.7 T2 

T2 = 41.6 C

Answer 2

Explanation:

It is given that mass of ice is 52.5 g and its number of moles will be calculated as follows.

      No. of moles = $\frac{mass}{\text{molar mass}}$

                            = $\frac{52.5 g}{18.0 g/mol}$

                            = 2.91 mol

Also, heat gained by ice cubes would be equal to the heat lost by warm water.

Therefore, heat required to melt all of the ice will be as follows.

    2.91 mol X 6.02 kJ/mol = 17.51 kJ = 17510 J       (As 1 kJ = 1000 J)

Now, calculate whether the warm water will still be above $0^{o}C$ when it loses this much heat:

-17510 J = $160 g \times 4.184 J/g^{o}C \times (\Delta T)$

             = $-26^{o}C$ 

Hence, calculate the final temperature as follows.

              q(ice/water) = - q(warm water) 

$\text{no. of moles} \times (\Delta H_{f}) + m \times c (T_{2} - T_{1}) = - m c (T_{2} - T_{1})$ 

    $2.91 mol \times 6.02 kJ/mol + 52.5g (4.184 J/gC) (T_{2} - 0) = -160g (4.184 J/g^{o}C) \times (T_{2} - 90)$

                     $T_{2} = 13.33^{o}C$

Thus, we can conclude that the final temperature of the given mixture is $13.33^{o}C$.