You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally insulated chamber. For water L = 3.33 x 105 J/kg. The specific heat of ice is 2090 J/(kg°C) and of water 4186 J/(kg°C). A. Everything turns to ice at a temperature below 0°C. B. Everything melts and is at a temperature above 0°C. C. There is a mixture of water and ice as the final state. D. The water and ice never reach the same temperature. E. There is not enough information to find the final temperature.

Let [tex]x\; \rm ^\circ C[/tex] be the final temperature of the mixture. Consider: what are some of the possible values of [tex]x[/tex]? There are three possible outcomes:

[tex]-20 < x < 0[/tex], the final mixture contains only ice.
[tex]0 < x < 2[/tex], the final mixture contains only water.
[tex]x = 0[/tex], the final mixture contains both ice and water.

Assumption 1

Assume that the final mixture contains only ice. In other words, assume that [tex]-20 < x < 0[/tex].

Energy released

Under this assumption, there are three ways by which energy is released:

[tex]Q_1[/tex], the energy released when [tex]m_2 = 1\; \rm kg[/tex] of water is cooled from [tex]2\; \rm ^\circ C[/tex] to [tex]0\; \rm ^\circ C[/tex].
[tex]Q_2[/tex], the energy released when [tex]m_2 = 1\; \rm kg[/tex] of water is turned into ice.
[tex]Q_3[/tex], the energy released when [tex]m_2 = 1\; \rm kg[/tex] of ice (from water) is cooled from [tex]0\; \rm ^\circ C[/tex] to [tex]x\; \rm ^\circ C[/tex].

The first and third of these three energies can be found with the equation:

[tex]Q = c\, m \, \Delta T[/tex], where

[tex]Q[/tex] is the energy change due to a change in temperature.
[tex]c[/tex] is the specific heat capacity.
[tex]m[/tex] is the mass of the object, and
[tex]\Delta T[/tex] is the change in temperature. Note that [tex]\Delta T = T(\text{final}) - T(\text{initial})[/tex].

Energy released when [tex]m_2 = 1\; \rm kg[/tex] of water is cooled from [tex]2\; \rm ^\circ C[/tex] to [tex]0\; \rm ^\circ C[/tex]:

[tex]Q_1 &= - c(\text{water}) \cdot m_2 \cdot \Delta T = 8372\; \rm J[/tex].

Since cooling releases energy, the output of [tex]c\, m \, \Delta T[/tex] would be negative. Add a minus sign to make sure that the value of [tex]Q_1[/tex] would be positive.

For the energy released when [tex]m_2 = 1\; \rm kg[/tex] of ice (from water) is cooled from [tex]0\; \rm ^\circ C[/tex] to [tex]x\; \rm ^\circ C[/tex]:

[tex]Q_3 &= -c(\text{ice}) \cdot m_2 \cdot \Delta T= (-2090\, x)\; \rm J[/tex].

Note that since [tex]x[/tex] is assumed to be negative, [tex](-2090\, x)[/tex] would be positive.

The second of these three energies can be found from [tex]L[/tex] the latent heat of fusion of water.

[tex]\begin{aligned} Q_2 &= m_2 \, L(\text{fusion, $\mathrm{H_2O}$}) \\ &= 1 \times 3.33 \times 10^5 = 3.33\times 10^5\; \rm J \end{aligned}[/tex].

Combine the three energies:

[tex]Q(\text{released}) = Q_1 + Q_2 + Q_3 = (341372 - 2090\, x)\; \rm J[/tex].

Energy absorbed

Under this assumption, there is only one way by which energy is absorbed:

Energy absorbed when [tex]m_2 = 1\; \rm kg[/tex] of ice is heated from [tex]-20\; \rm ^\circ C[/tex] to [tex]x \; \rm ^\circ C[/tex].

Similarly, this energy can be found with the equation [tex]Q = c \, m \, \Delta T[/tex].

[tex]\begin{aligned}Q &= c(\text{ice}) \cdot m_1 \cdot \Delta T \\&= 2090 \times 1 \times (x - 20) \\&= (2090\, x -41800)\; \rm J \end{aligned}[/tex].

Since this [tex]Q[/tex] would be the only energy absorbed under this assumption:

[tex]Q(\text{absorbed}) = (2090\, x - 41800)\; \rm J[/tex].

Solve for temperature and test this assumption

Since this mixture is thermally insulated, [tex]Q(\text{released}) = Q(\text{absorbed})[/tex]. Therefore:

[tex]341372 - 2090\, x = 2090\, x - 41800[/tex].

Solve this equation for [tex]x[/tex]:

[tex]x \approx 91.668[/tex].

That does not satisfy [tex]-20 < x < 0[/tex]. Hence, this assumption is not valid.

Assumption 2

Assume that the final mixture contains only water. In other words, assume that [tex]0 < x < 2[/tex].

The calculations are similar. For the heat released when melting ice, note that under the same temperature and pressure,

the heat absorbed when melting [tex]1\; \rm kg[/tex] of ice

is the same as

the heat released when freezing [tex]1\; \rm kg[/tex] of water.

[tex]Q(\text{released}) = 8372 - 4186\, x[/tex].[tex]Q(\text{absorbed}) = Q_1 + Q_2 + Q_3 = 4186\, x+ 416702[/tex].

Solve for temperature and test this assumption

Again, [tex]Q(\text{released}) = Q(\text{absorbed})[/tex]. Therefore:

[tex]8372 - 4186\, x = 4186\, x+ 416702[/tex]

[tex]x \approx -48.773[/tex].

That does not satisfy [tex]0 < x < 2[/tex]. Hence, this assumption is not valid, either.

Conclusion

Exactly one of these three assumptions should be valid. Since assumption 1 and 2 are not valid, assumption 3 should be valid. In other words, the final temperature will be [tex]0 \; \rm ^\circ C[/tex], and the mixture will contain both ice and water.