# An object with a mass of 4 kg, temperature of 240 ^oC, and a specific heat of 15 J/(kg*K) is dropped into a container with 15L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Apr 8, 2016

The object does not vaporise the water. It causes a temperature change of ${0.23}^{\circ} \text{C}$ in the water.

#### Explanation:

Evaporation of a liquid can occur at any temperature. So we could say "Yes" in answer to the question "Does the water evaporate?" However, let’s assume the question meant "Does all of the water get vaporised?" (I.e. is the water raised to its boiling point and then boiled away?)

To answer the question I will structure my solution as follows:

1. Calculate the energy required to raise the temperature of the water to 100ºC.
2. Calculate the energy required to completely vaporise the water.
3. Calculate the energy available for transfer from the object and compare that value to energies 1 and 2 above.
4. If the object does not have enough energy to vaporise the water the temperature change in the water will be calculated.

Some preliminary data and calculations.

Conversion of the volume of water into m³:

V = 15 L = 15 × 10^(-3) m^3

Calculation of the mass of water:

m_w = ρV = 1000 × 15 × 10^(-3) = 15 kg

Specific heat capacity of water:

${c}_{w} = 4200 J . k {g}^{- 1} . {K}^{- 1}$

Latent heat of vaporisation of water:

${L}_{v \left(w a t e r\right)} = 334 k J . k {g}^{- 1}$

An initial note about the question.

As you can see from the data that was provided the object has both a much smaller specific heat capacity and a smaller mass than the water.

So we can provisionally tell that the impact of the object on the water is going to be very small indeed. It is in fact so small that I wonder whether some of the data provided was incorrect?

1. Energy required to increase temperature of water to boiling point.

DeltaQ_1 = m_wc_wDeltaθ = 15 × 4200 × (100 - 0)
⇒ DeltaQ_1 = 6.3 MJ

2. Energy required to vaporise all of the water.

DeltaQ_2 = m_wL_(v(water)) = 15 × 334 × 10^3
⇒ DeltaQ_2 = 5.0 MJ

3. Energy available from the 4kg object.

The lowest temperature the object could fall to and successfully vaporise the water would be ${100}^{\circ} \text{C}$. So the maximum temperature change would be from ${240}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$.

Data for the object: ${m}_{o} = 4 k g$, ${c}_{o} = 15 J . k {g}^{- 1} . {K}^{- 1}$

DeltaQ_3 = m_oc_oDeltaθ = 4 × 15 × (240 - 100) = 8400 J

Compare the energies. We can see that 8400 J is much less than 6.4 MJ (∆Q_1). So the object is unable to heat the water to anywhere near its boiling point.

We don’t even need to consider comparing 8400 J with the energy required to vaporise the water (calculated in part 2).

4. By how much does the water’s temperature change?

Since we know the water does not undergo a phase change we will only need to use the specific heat capacity equation for both the object and the water.

I will define the final temperature of the water as θ_x. Since all energy transfers stop when both the object and the water have the same temperature the object’s final temperature will also be θ_x.

Object starts at ${240}^{\circ} \text{C}$ and ends at θ_x.
Water starts at ${0}^{\circ} \text{C}$ and ends at θ_x.

Assuming that all heat energy transferred by the object is absorbed by the water we can state the following:

$\Delta {Q}_{w} = \Delta {Q}_{o}$
⇒ m_wc_wDeltaθ_w = m_oc_oDeltaθ_o

Noting the above information about the temperature changes we can further define the ∆θ terms:

Deltaθ_w = θ_x – 0
Deltaθ_o = 240 – θ_x

Substitute those into equation ①:

⇒ m_wc_w(θ_x – 0) = m_oc_o(240 – θ_x)
⇒ m_wc_wθ_x = 240m_oc_o – m_oc_oθ_x

Rearrange for θ_x:

⇒ θ_x (m_wc_w + m_oc_o) = 240m_oc_o
⇒ θ_x = (240m_oc_o)/(m_wc_w + m_oc_o)

Now substitute the values into the equation:

⇒ θ_x = (240 × 4 × 15)/(15 × 4200 + 4 × 15) = 0.23^@"C"

So the temperature change in the water is ${0.23}^{\circ} \text{C}$, i.e. ${0}^{\circ} \text{C}$ to ${0.23}^{\circ} \text{C}$.