# Physics heat exercises and questions

###### tags: `Physics` `4º ESO` `Heat` `work`

I've made this so you can have exam-like questions and answers.

[ToC]

## Theoretical concepts

(The concepts that can be asked regarding theory are from page 226 to 236)

* What is heat
* What is temperature
* What is specific heat of a substance
* What is the latent heat of a substance
* What are the units that we can use to measure heat (cal and J)
* What is an internal combustion engine
* What is an external combustion engine
* What is a thermal machine
* What is a reversed thermal machine
* What is the water equivalent of a calorimeter

## Practical concepts

There is going to be some easy exercises and some more advance exercises to get to the 10. In both of them I will give you a table that it's going to be at the end of the exam where you can check out the data.

You need to know if you need the specific heat, the boiling latent heat or the melting latent heat in each situation. (so you need to read carefully the table)

In this document I'm leaving the data in the [Appendix](https://hackmd.io/2ZkTEs4MSDGLzCX5QUYZjw#Appendix-data-tables)

**Remember to put properly the units**

### Easy exercises

You will se 1-3 exercises like this in the exam.

#### How much heat do you need to melt iron?

We have a 3 kg of Iron at 25 ºC. How much energy do we need to melt it completely?

Solution:
- Revise the data that we need to find.

In this case, in order to melt Iron we need to heat it up until the **temperature of melting**. Then we need to melt the iron. We already know the actual temperature and the mass of the iron.

We need to split this into 2: **Q~1~** that is the energy to heat the iron and **Q~2~** that is the energy that you need to melt it.

- check in the tables

So we need to check in the table for the specific heat of Iron to know how much energy we need to heat it up until the temperature.

In this case:

Specific heat of Iron: 452 Jkg^-1^K^-1^
Melting temperature of Iron: 1482 ºC
Latent heat of Fusion (L~f~): 293 kJkg^-1^ (remember that this is given in **kJ**kg^-1^, not in **J**kg^-1^)

- Use the formulas

$Q_1 = m· c· \Delta T = 3 kg · 452 Jkg^{-1}K^{-1} · (1482 ºC - 25ºC) = 19 715 629 J \approx 1972 kJ$

$Q_2 = m · L_f = 3 kg · 293 kJ/kg = 879 kJ$

Be careful with the units over here!

$Q_{total} = 19715629 J + 879 kJ = 2854692 J \approx 2854 kJ$

**Try it yourself**. Let's say that we have this question:

We have a 10.5 kg of Iron at 10 ºC. How much energy do we need to melt it completely?

The answer should be 10 063 kJ /10 062 612 J

#### Simple thermal equlibrium

I have 100 ml of water at 60 ºC and pour it in a thermal container with 200 ml of oil at 20 º C, what is the temperature of the liquid after some time? (don't consider outside loses of heat)

Data:
Oil density - 750 gr/l

Solution:

Here we have first to pass all the quantities into mass units to work properly. Then we will check that the final temperature for both parts will be the same

- Passing the quantities into kilograms

Water: 100 ml = 0.1 l
Density of water: 1 kg/l
Mass: 0.1 l · 1kg/l = 0.1 kg

Oil: 200 ml = 0.2 l
Density of oil: 750 gr/l
Mass: 0.2l · 750 gr/l = 150 gr = 0.15 kg

_remember to put all in the same units, kg grams, etc_

- Check for other data in the tables

We're using the thermal equilibrium so we will need the specific heats of oil and water

Water(liquid state): 4180 Jkg^-1^K^-1^
Olive oil: 1675 Jkg^-1^K^-1^

- Thermal equilibrium

The heat given by the water (Q~1~) to the oil has to be the same as the heat taken by the oil(Q~2~). There is no heat lost so the equation will be:

$Q_1 + Q_2 = 0$

Now we develop this using:

$Q = m· c· \Delta T$

First the water:
(don't be afraid if there are some unknowns yet)

$Q_1 = m· c· \Delta T = 0.1 kg * 4180 Jkg^{-1}K^{-1} · (T_{eq} - 60ºC)$

Then the oil

$Q_2 = m· c· \Delta T = 0.15 kg * 1675 Jkg^{-1}K^{-1} · (T_{eq} - 20ºC)$

We know that

$Q_1 + Q_2 = 0$

So we can state also:

$Q_1 = - Q_2$

So we substitute here:

$0.1 kg * 4180 Jkg^{-1}K^{-1} · (T_{eq} - 60ºC) = - 0.15 kg * 1675 Jkg^{-1}K^{-1} · (T_{eq} - 20ºC)$

$418· T_{eq} - 418· 60 = - 251.25 · T_{eq} - 251.25 · (-20)$

$(418 +251.25) · T_{eq} = 418· 60 + 251.25 · 20$

$669.25 · T_{eq} = 25080 + 5025 = 30105$

$T_{eq} = 30105/ 669.25 = 44.98 ºC$

**Try it yourself!**

Let's say that we have this question:

I have 150 ml of water at 20 ºC and pour it in a thermal container with 100 ml of oil at 55 º C, what is the temperature of the liquid after some time? (don't consider outside loses of heat)

Data:
Oil density - 750 gr/l

It should give you **25.84ºC**

#### Find the water equivalent of a calorimiter.

This is something that I'm going to ask in the exam, you should know how to do it because it's what you have to use in the lab report.

This is the exercise in page 229:

We heat 50 grams of water up to 100 ºC
We collect 100 grams of water and we put in the calorimeter. That water it's at 20 ºC
We mix both waters in the calorimeter and we wait for the temperatue to stabilise. It finally gets to 42ºC

**What is the water equivalent of the calorimeter that I'm using?**

Solution:

For this we need to asume the thermal equilibrium between the 3 parts in touch. The hot water, the cold water and the calorimeter. And the 3 have to sum 0.

$$Q_{cold water}+Q_{calorimeter}+Q_{hot water}=0$$

And we split

$Q_{cold water} = 0.1kg · 4180 Jkg^{-1}K^{-1} · (42 ºC-20 ºC)$

$Q_{cold water} = 9196 J$

$Q_{hot water} = 0.05kg · 4180 Jkg^{-1}K^{-1} · (42 ºC-100 ºC)$

$Q_{hot water} = -12122 J$
This is in negative because the heat is taken **from** the hot water.

$Q_{calorimeter} = m_{wat\ eq\ cal} · 4180 Jkg^{-1}K^{-1} · (42 ºC-20 ºC)$

Here we have an unknown: $m_{wat\ eq\ cal}$ so we find that Q in terms of that unknown.

$Q_{calorimeter} = 91\ 960· m_{wat\ eq\ cal}\ J$

We put together all three terms:

$$Q_{cold water}+Q_{calorimeter}+Q_{hot water}=0 = 9196 J + 91\ 960· m_{wat\ eq\ cal}\ J -12122 J$$

And we find $m_{wat\ eq\ cal}$

$91\ 960· m_{wat\ eq\ cal}\ = [phone] = 2926\ J$

$m_{wat\ eq\ cal}\ = 2926/ 91\ 960 = 0.0318 kg \approx 0.032kg = 32g$

Try it yourself!

Supose that in this case we have this data:

We heat 50 grams of water up to 100 ºC
We collect 80 grams of water and we put in the calorimeter. That water it's at 20 ºC
We mix both waters in the calorimeter and we wait for the temperatue to stabilise. It finally gets to 47ºC

*What is the water equivalent of the calorimeter that I'm using?*

The result should be exactly 26 grams

#### Use latent heat to find heat needed

How much energy do we need to vaporise 4 litters of liquid water at 100ºC?

Solution:

First: We check that the only temperature that we have here is the vaporization of water. This exercise can be mistaken with others that they ask you to heat something and then boil/melt it.

We check in the table that, indeed, the water boils at 100º C and we don't need to heat anything.

Then we need to find the mass that we have: 4 l = 4kg in the case of water.

Last we need to find the Latent heat of vaporization. How much energy do we need to boil a unit of mass. That's 6300 kJ/kg.

$Q_2 = m · L_v = 4 kg · 6\ 300 kJ/kg = 25\ 200 kJ$

The answer is 25 200 kJ.

Try it yourself!

How much energy do we need to vaporise 500 grams of oxigen at -183 ºC?

The result should be 106.5 kJ

---

### Advanced exercises

You will see at most 1 exercise of this kind. It will be, at most 1.5 marks.

Check anyway the solved problem in page 231

#### Find the specific heat using a calorimeter

I give you the water equivalent of a calorimeter (or maybe it's the result of the previous exercise) and you find the specific heat of a ficticious metal such as Adamantium.

Let's try it!

We have a calorimeter which water equivalent is 40 grams of water. In that calorimeter we have 120 ml of water at room temperature (20ºC)

We heat a strange substance that came from an U.F.O. that has a mass of 30 grams. We heat that substance up to 90 degrees celsius.

Then we put the strange substance in the calorimeter and, luckily, it didn't explode.

We wait a bit and the temperature in the calorimeter goes to 30 degrees celsius where it gets stabilized.

*What is the specific heat of the substance that we took from the UFO?*

Solution

Here we have 3 Q

$Q_1$ is the heat that the room temperature water absorbs from the substance.

$Q_2$ is the heat that the calorimeter (that it's also at room temperature) gets from the substance

$Q_3$ is the heat loses the strange substance to both calorimeter and the water.

The three of them have to add 0 because this energy is not going to go anywhere.

$$Q_{1}+Q_{2}+Q_{3}=0$$

The only unknown that we have is the specific heat of the strange substance frome the UFO. All the other data we can take it from tables or from the statement. Let's go 1 by 1 ($Q$ by $Q$):

$Q_1 = m ·c_{water} · \Delta T = 0.08 kg · 4180 Jkg^{-1}K^{-1} · (30 ºC-20 ºC) = 3344\ J$

$Q_2 = m_{wat\ eq\ cal} ·c_{water} · \Delta T = 0.04 · 4180 Jkg^{-1}K^{-1} · (30 ºC-20 ºC) = 1672\ J$

$Q_3 = m_{strange\ substance} ·c_{strange\ substance} · \Delta T = 0.03 ·c_{strange\ substance} · (30ºC - 90ºC) = - 1.8 ·c_{strange\ substance} \ J$

And we put together in one equation that we can easily solve:

$3344 + 1672- 1.8 ·c_{strange\ substance} = 0$

$c_{strange\ substance} = \frac{3344 + 1672}{1.8} = 2786\ Jkg^{-1}K^{-1}$

Extra comment:

Of course you can go and write everything in just one equation but I think that it's way easier to solve this:

$3344 + 1672- 1.8 ·c_{strange\ substance} = 0$

Than this:

$0.08 kg · 4180 Jkg^{-1}K^{-1} · (30 ºC-20 ºC) + 0.04 · 4180 Jkg^{-1}K^{-1} · (30 ºC-20 ºC) -0.03 ·c_{strange\ substance} · (30ºC - 90ºC) = 0$

Even if it's actually the same. It's harder to commit any mistakes. Divide and conquer!

Now try it yourself!

We have a calorimeter which water equivalent is 32 grams of water. In that calorimeter we have 100 ml of water at room temperature (23ºC)

He heat a strange substance that came from an U.F.O. that has a mass of 35 grams. We heat that substance up to 95 degrees celsius.

Then we put the strange substance in the calorimeter and, luckily, it didn't explode.

We wait a bit and the temperature in the calorimeter goes to 29 degrees celsius where it gets stabilized.

*What is the specific heat of the substance that we took from the UFO?*

The answer with my data is 128.06 J/kg/K

#### Finding the temperature of equilibrium with change of state

In the page 231 there is a problem with ice that melts but not completely. To do that you need to make the proportion of energy and you can have a measure of how much ice is going to melt

#### Find Latent heat of fusion with a calorimeter

As we have measured specif heat we can find the Latent heat of fusion or vaporisation doing an experiment. I could give you some info about a substance and then give you water and the water equivalent.

## Appendix data tables

### Specific heats

| Substance | Specific heat Jkg^-1^K^-1^ |
| -------- | -------- |
| Iron | 452 |
| Water(ice) | 2090 |
| Water(liquid state) | 4180 |
| Water (steam) | 1920 |
| Olive oil | 1675|

### Latent heats:

| Substance | Latent heat of Fusion (L~f~) kJkg^-1^ | T~melting~ ºC | Latent heat of Vaporisation (L~v~) kJkg^-1^ |T~boiling~ ºC |
| -------- | -------- |-------- | -------- | -------- |
| Iron | 293 | 1482 | 6300 | 2750 |
| Ice (Water) | 334 | 0 | 2260 | 100 |
|Oxygen | 13.9 | −219 | 213 | −183 |