Physics, 25.05.2020 12:57 smart1212

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.

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Answer from: dariusardelean8700

Choice C. There is a mixture of water and ice as the final state.

Explanation:

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

, the final mixture contains only ice. 0 < x < 2

, the final mixture contains only water. x = 0

, the final mixture contains both ice and water.Assumption 1

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

Energy released

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

, the energy released when $m_2 = 1\; \rm kg$ of water is cooled from $2\; \rm ^\circ C$ to $0\; \rm ^\circ C$ . Q_2

, the energy released when $m_2 = 1\; \rm kg$ of water is turned into ice. Q_3

, the energy released when $m_2 = 1\; \rm kg$ of ice (from water) is cooled from $0\; \rm ^\circ C$ to $x\; \rm ^\circ C$ .

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

$Q = c\, m \, \Delta T$ , where

is the energy change due to a change in temperature.

is the specific heat capacity.

is the mass of the object, and $\Delta T$ is the change in temperature. Note that $\Delta T = T(\text{final}) - T(\text{initial})$ .

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

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

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

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

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

Note that since is assumed to be negative, $(-2090\, x)$ would be positive.

The second of these three energies can be found from the latent heat of fusion of water.

$\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}$ .

Combine the three energies:

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

Energy absorbed

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

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

Similarly, this energy can be found with the equation $Q = c \, m \, \Delta T$ .

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

Since this would be the only energy absorbed under this assumption:

$Q(\text{absorbed}) = (2090\, x - 41800)\; \rm J$ .

Solve for temperature and test this assumption

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

$341372 - 2090\, x = 2090\, x - 41800$ .

Solve this equation for :

$x \approx 91.668$ .

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

Assumption 2

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

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

the heat absorbed when melting $1\; \rm kg$ of ice

is the same as

the heat released when freezing $1\; \rm kg$ of water.

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

Solve for temperature and test this assumption

Again, $Q(\text{released}) = Q(\text{absorbed})$ . Therefore:

$8372 - 4186\, x = 4186\, x+ 416702$

$x \approx -48.773$ .

That does not satisfy 0 < x < 2 . 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 $0 \; \rm ^\circ C$ , and the mixture will contain both ice and water.

Answer from: Quest

1. may 2. september 3. february, and december4. 70°f

Answer from: Quest

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this. hope that you like the video and that it explains more in detail to you.

that is the link.

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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 thermall...

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