Respuesta :
Answer:
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 ([tex]Q_{W}[/tex]) due to the hot water will be equal to the negative of the one ([tex]Q_{I}[/tex]) due to the melted ice.
i.e
[tex]Q_{W}[/tex] = -[tex]Q_{I}[/tex] -----------------(a)
Where;
[tex]Q_{I}[/tex] = [tex]m_{I}[/tex] x [tex]c_{I}[/tex] x Δ[tex]T_{I}[/tex] [[tex]m_{I}[/tex] = mass of ice, [tex]c_{I}[/tex] = specific heat capacity of melted ice which is now water and Δ[tex]T_{I}[/tex] = change in temperature of the melted ice]
and
[tex]Q_{W}[/tex] = [tex]m_{W}[/tex] x [tex]c_{W}[/tex] x Δ[tex]T_{W}[/tex]
[[tex]m_{W}[/tex] = mass of water, [tex]c_{W}[/tex] = specific heat capacity of water and Δ[tex]T_{W}[/tex] = change in temperature of the water]
Substitute the values of [tex]Q_{W}[/tex] and [tex]Q_{I}[/tex] into equation (a) as follows
[tex]m_{W}[/tex] x [tex]c_{W}[/tex] x Δ[tex]T_{W}[/tex] = - [tex]m_{I}[/tex] x [tex]c_{I}[/tex] x Δ[tex]T_{I}[/tex]
Note that [tex]c_{W}[/tex] and [tex]c_{I}[/tex] are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;
[tex]m_{W}[/tex] x Δ[tex]T_{W}[/tex] = -[tex]m_{I}[/tex] x Δ[tex]T_{I}[/tex] -----------------------(b)
Now, let's analyse Δ[tex]T_{W}[/tex] and Δ[tex]T_{I}[/tex]. The final temperature ([tex]T_{F}[/tex]) of the two kinds of water(melted ice and cooled water) are now the same.
=> Δ[tex]T_{W}[/tex] = change in temperature of water = final temperature of water([tex]T_{F}[/tex]) - initial temperature of water([tex]T_{IW}[/tex])
Δ[tex]T_{W}[/tex] = [tex]T_{F}[/tex] - [tex]T_{IW}[/tex]
Where;
[tex]T_{IW}[/tex] = 22°C [which is the final temperature of water before mixture]
=> Δ[tex]T_{I}[/tex] = change in temperature of melted ice = final temperature of water([tex]T_{F}[/tex]) - initial temperature of melted ice ([tex]T_{II}[/tex])
Δ[tex]T_{I}[/tex] = [tex]T_{F}[/tex] - [tex]T_{II}[/tex]
[tex]T_{II}[/tex] = 0°C (Initial temperature of the melted ice)
Substitute these values into equation (b) as follows;
[tex]m_{W}[/tex] x Δ[tex]T_{W}[/tex] = - [tex]m_{I}[/tex] x Δ[tex]T_{I}[/tex]
0.400 x ([tex]T_{F}[/tex] - [tex]T_{IW}[/tex]) = -0.0550 x ([tex]T_{F}[/tex] - [tex]T_{II}[/tex])
0.400 x ([tex]T_{F}[/tex] - 22) = -0.0550 x ([tex]T_{F}[/tex] - 0)
0.400 x ([tex]T_{F}[/tex] - 22) = -0.0550 x ([tex]T_{F}[/tex])
0.400[tex]T_{F}[/tex] - 8.8 = -0.0550[tex]T_{F}[/tex]
0.400[tex]T_{F}[/tex] + 0.0550[tex]T_{F}[/tex] = 8.8
0.455[tex]T_{F}[/tex] = 8.8
[tex]T_{F}[/tex] = 19.34°C
Therefore, the final temperature of the mixture is 19.34°C
