Experimental problem in IPHO

Meant for those preparing for physics olympiad. I'll be giving more of the practical problems in this form from earlier olympiads. Do reshare them.

Experimental problem

The following devices and materials are given:

  1. Balance (without weights)
  2. Calorimeter
  3. Thermometer
  4. Source of voltage
  5. Switches
  6. Wires
  7. Electric heater
  8. Stop-watch
  9. Beakers
  10. Water
  11. Petroleum
  12. Sand (for balancing)

Determine specific heat of petroleum. The specific heat of water is 1 cal/(g⋅°C). The specific heat of the calorimeter is 0.092 cal/(g⋅°C). Discuss assumptions made in the solution. Solution.

The devices given to the students allowed using several methods. The students used the following three methods:

  1. Comparison of velocity of warming up water and petroleum;
  2. Comparison of cooling down water and petroleum;
  3. Traditional heat balance.

As no weights were given, the students had to use the sand to find portions of petroleum and water with masses equal to the mass of calorimeter.

First method: comparison of velocity of warming up. If the heater is inside water then both water and calorimeter are warming up. The heat taken by water and calorimeter is:

mwcwΔt1+mcccΔt1m_{w}c_{w}\Delta t_{1}+m_{c}c_{c}\Delta t_{1}

where: wmw_{m} denotes mass of water, cmc_{m}- mass of calorimeter, wcw_{c}- specific heat of water, ccc_{c}- specific heat of calorimeter, Δt1\Delta t_{1}- change of temperature of the system water + calorimeter.

On the other hand, the heat provided by the heater is equal

Q=AU2RτQ=A\frac {U^{2}}{R}\tau

where:A denotes the thermal equivalent of work,U - voltage, R – resistance of the heater, τ1τ_{1} – time of work of the heater in the water. Of course, Q1=Q2Q_{1}=Q_{2}. Thus mwcwΔt1+mcccΔt1=AU2Rτ1m_{w}c_{w}\Delta t_{1}+m_{c}c_{c}\Delta t_{1}=A\frac {U^{2}}{R}\tau_{1}.

For petroleum in the calorimeter we get a similar formula mpcpΔt1+mcccΔt2=AU2Rτ2m_{p}c_{p}\Delta t_{1}+m_{c}c_{c}\Delta t_{2}=A\frac {U^{2}}{R}\tau_{2}. where: pmp_{m} denotes mass of petroleum, pcp_{c}- specific heat of petroleum, Δt2\Delta t_{2}- change of temperature of the system water + petroleum, τ2\tau_{2}– time of work of the heater in the petroleum.? By dividing the last equations we get τ1τ2=mwcwΔt1+mcccΔt2mpcpΔt1+mcccΔt2\frac {\tau_{1}}{\tau_{2}}=\frac{m_{w}c_{w}\Delta t_{1}+m_{c}c_{c}\Delta t_{2}}{m_{p}c_{p}\Delta t_{1}+m_{c}c_{c}\Delta t_{2}}

It is convenient to perform the experiment by taking masses of water and petroleum equal to the mass of the calorimeter (for that we use the balance and the sand). For mw=mp=mcm_{w}=m_{p}=m_{c}

τ1τ2=cwΔt1+ccΔt2cpΔt1+ccΔt2\frac {\tau_{1}}{\tau_{2}}=\frac {c_{w}\Delta t_{1}+c_{c}\Delta t_{2}}{c_{p}\Delta t_{1}+c_{c}\Delta t_{2}}

cc=Δt1τ1τ2Δt2cw(1Δt1τ1τ2Δt2)ccc_{c}=\frac {\Delta t_{1}}{\tau_{1}}\frac {\tau_{2}}{\Delta t_{2}}c_{w}-(1-\frac {\Delta t_{1}}{\tau_{1}}\frac {\tau_{2}}{\Delta t_{2}})c_{c}

cc=k1k2cw(1k1k2)ccc_{c}=\frac {k_{1}}{k_{2}}c_{w}-(1-\frac {k_{1}}{k_{2}})c_{c}

k1=Δt1τ1k_{1}=\frac {\Delta t_{1}}{\tau_{1}} and k2=Δt2τ2k_{2}=\frac {\Delta t_{2}}{\tau_{2}}

denote “velocities of heating” water and petroleum, respectively. These quantities can be determined experimentally by drawing graphs representing dependence ∆t1and ∆t2 on time (τ). The experiment shows that these dependences are linear. Thus, it is enough to take slopes of appropriate straight lines. The experimental setup given to the students allowed measurements of the specific heat of petroleum, equal to 0.53 cal/(g°⋅C), with accuracy about 1%.

Second method: comparison of velocity of cooling down

Some students initially heated the liquids in the calorimeter and later observed their cooling down. This method is based on the Newton’s law of cooling. It says that the heat Q transferred during cooling in time τ is given by the formula:

Q=τ(tθ)sh)Q=τ(t-\theta)sh) ( where: t denotes the temperature of the body, ϑ - the temperature of surrounding, s – area of the body, and h – certain coefficient characterizing properties of the surface. This formula is correct for small differences of temperatures ϑ−t only (small compared to t and ϑ in the absolute scale). This method, like the previous one, can be applied in different versions. We will consider only one of them. Consider the situation when cooling of water and petroleum is observed in the same calorimeter (containing initially water and later petroleum). The heat lost by the system water + calorimeter is Q1=(mwcw+mccc)ΔtQ_{1}=(m_{w}c_{w}+m_{c}c_{c})\Delta t

where ∆t denotes a change of the temperature of the system during certain period τ1. For the system petroleum + calorimeter, under assumption that the change in the temperature ∆t is the same, we have

Q2=(mpcp+mccc)ΔtQ_{2}=(m_{p}c_{p}+m_{c}c_{c})\Delta t

Of course, the time corresponding to ∆t in the second case will be different. Let it be τ2. From the Newton's law we get Q1Q2=τ1τ2\frac {Q_{1}}{Q_{2}}=\frac {\tau_{1}}{\tau_{2}}

Thus

τ1τ2=mwcw+mcccmpcp+mccc\frac {\tau_{1}}{\tau_{2}}=\frac{m_{w}c_{w}+m_{c}c_{c}}{m_{p}c_{p}+m_{c}c_{c}}

If we conduct the experiment at mw=mp=mcm_{w}=m_{p}=m_{c}

then we get

cc=T2T1cw(1T2T1)ccc_{c}=\frac {T_{2}}{T_{1}}c_{w}-(1-\frac {T_{2}}{T_{1}})c_{c}

Third method: heat balance

This method is rather typical. The students heated the water in the calorimeter to certain temperature 1t and added the petroleum with the temperature 2t. After reaching the thermal equilibrium the final temperature was t. From the thermal balance (neglecting the heat losses) we have

(mwcw+mccc)(t1t)=mpcp(tt2)(m_{w}c_{w}+m_{c}c_{c})(t_{1}-t)=m_{p}c_{p}(t-t_{2}) If, like previously, the experiment is conducted at mw=mc=mpm_{w}=m_{c}=m_{p}

cp=(cw+cc)t1ttt2c_{p}=(c_{w}+c_{c})\frac {t_{1}-t}{t-t_{2}}

In this methods the heat losses (when adding the petroleum to the water) always played a substantial role. The accuracy of the result equal or better than 5% can be reached by using any of the methods described above. However, one should remark that in the first method it was easiest. The most common mistake was neglecting the heat capacity of the calorimeter. This mistake increased the error additionally by about 8% .

Hope you liked it follow me for mere and do no not forget to reshare and leave your valuable feedback.

Note by Satvik Choudhary
4 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Are you a participant of IPHOTC this year?

Krishna Ar - 4 years, 4 months ago

Log in to reply

No i will try next year. Can you tell me about you i am kinda new on brilliant.

Satvik Choudhary - 4 years, 4 months ago

Log in to reply

Nothing much, I am just another 15 yr old here on this site.. I joined almost at the same time as you joined!

Krishna Ar - 4 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...