A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temperature in degrees Celsius?

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂                  ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

Where;

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁]        ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and  L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂                -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T  = 21848.2 / 1680

=> 35 - T  = 13

=> T  = 35 - 13

=> T  = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat () due to the hot water will be equal to the negative of the one () due to the melted ice.

i.e

= -             -----------------(a)

Where;

= x x Δ         [ = mass of ice, = specific heat capacity of melted ice which is now water and Δ = change in temperature of the melted ice]

and

= x x Δ    

[ = mass of water, = specific heat capacity of water and Δ = change in temperature of the water]

Substitute the values of and into equation (a) as follows

x x Δ   =  - x x Δ

Note that and are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

x Δ   = - x Δ   -----------------------(b)

Now, let's analyse Δ and Δ. The final temperature () of the two kinds of water(melted ice and cooled water) are now the same.

=> Δ = change in temperature of water = final temperature of water() - initial temperature of water()

Δ = -

Where;

= 22°C           [which is the final temperature of water before mixture]

=> Δ = change in temperature of melted ice = final temperature of water() - initial temperature of melted ice ()

Δ = -

= 0°C     (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

x Δ   =  - x Δ

0.400 x ( - ) = -0.0550 x ( - )

0.400 x ( - 22) = -0.0550 x ( - 0)

0.400 x ( - 22) = -0.0550 x ()

0.400 - 8.8 = -0.0550

0.400 + 0.0550 =  8.8  

0.455 = 8.8

= 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

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