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:

$m_{w}c_{w}\Delta t_{1}+m_{c}c_{c}\Delta t_{1}$

where: $w_{m}$ denotes mass of water, $c_{m}$- mass of calorimeter, $w_{c}$- specific heat of water, $c_{c}$- specific heat of calorimeter, $\Delta t_{1}$- change of temperature of the system water + calorimeter.

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

$Q=A\frac {U^{2}}{R}\tau$

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

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

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 $m_{w}=m_{p}=m_{c}$

$\frac {\tau_{1}}{\tau_{2}}=\frac {c_{w}\Delta t_{1}+c_{c}\Delta t_{2}}{c_{p}\Delta t_{1}+c_{c}\Delta t_{2}}$

$c_{c}=\frac {\Delta t_{1}}{\tau_{1}}\frac {\tau_{2}}{\Delta t_{2}}c_{w}-(1-\frac {\Delta t_{1}}{\tau_{1}}\frac {\tau_{2}}{\Delta t_{2}})c_{c}$

$c_{c}=\frac {k_{1}}{k_{2}}c_{w}-(1-\frac {k_{1}}{k_{2}})c_{c}$

$k_{1}=\frac {\Delta t_{1}}{\tau_{1}}$ and $k_{2}=\frac {\Delta t_{2}}{\tau_{2}}$

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:

$Q=τ(t-\theta)sh)$ ( 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 $Q_{1}=(m_{w}c_{w}+m_{c}c_{c})\Delta t$

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

$Q_{2}=(m_{p}c_{p}+m_{c}c_{c})\Delta t$

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 $\frac {Q_{1}}{Q_{2}}=\frac {\tau_{1}}{\tau_{2}}$

Thus

$\frac {\tau_{1}}{\tau_{2}}=\frac{m_{w}c_{w}+m_{c}c_{c}}{m_{p}c_{p}+m_{c}c_{c}}$

If we conduct the experiment at $m_{w}=m_{p}=m_{c}$

then we get

$c_{c}=\frac {T_{2}}{T_{1}}c_{w}-(1-\frac {T_{2}}{T_{1}})c_{c}$

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

$(m_{w}c_{w}+m_{c}c_{c})(t_{1}-t)=m_{p}c_{p}(t-t_{2})$ If, like previously, the experiment is conducted at $m_{w}=m_{c}=m_{p}$

$c_{p}=(c_{w}+c_{c})\frac {t_{1}-t}{t-t_{2}}$

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% .

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