Question

В цинковый сосуд массой 250 грамм , содержащий керосин, объемом 0,5 литров при температуре 15( градусов цельсия), опустили стальную деталь массой 150 грамм. Какая была первоначальная температура детали, если в результате теплообмена в сосуде установилась температура 60( градусов цельсия)

Answer 1

Qц+Qк=Qд
Qц=cцmцdt=400*0,25*45
mк=pк*vк
Qк=cкmкdt=2100*(800*0,0005)*45
Qд=cстmст(t2-t1),t2=(Qд/cm)+t1=(Qд/500*0,15)+60

Answer 2

Problem Analysis

We have a zinc vessel with a mass of 250 grams, containing kerosene with a volume of 0.5 liters at a temperature of 15 degrees Celsius. A steel part with a mass of 150 grams is submerged in the vessel, and as a result of heat exchange, the temperature in the vessel stabilizes at 60 degrees Celsius. We need to determine the initial temperature of the steel part.

Solution

To solve this problem, we can use the principle of conservation of energy. The heat gained by the steel part is equal to the heat lost by the kerosene and the zinc vessel.

The heat gained by the steel part can be calculated using the formula:

Q1 = m1 * c1 * (Tf - Ti)

Where: - Q1 is the heat gained by the steel part - m1 is the mass of the steel part (150 grams) - c1 is the specific heat capacity of steel (approximately 0.45 J/g°C) - Tf is the final temperature (60 degrees Celsius) - Ti is the initial temperature of the steel part (unknown)

The heat lost by the kerosene and the zinc vessel can be calculated using the formula:

Q2 = (m2 * c2 + m3 * c3) * (Tf - Ti)

Where: - Q2 is the heat lost by the kerosene and the zinc vessel - m2 is the mass of the kerosene (unknown) - c2 is the specific heat capacity of kerosene (approximately 2.1 J/g°C) - m3 is the mass of the zinc vessel (250 grams) - c3 is the specific heat capacity of zinc (approximately 0.39 J/g°C) - Tf is the final temperature (60 degrees Celsius) - Ti is the initial temperature of the kerosene and the zinc vessel (15 degrees Celsius)

Since the heat gained by the steel part is equal to the heat lost by the kerosene and the zinc vessel, we can set up the following equation:

Q1 = Q2

m1 * c1 * (Tf - Ti) = (m2 * c2 + m3 * c3) * (Tf - Ti)

Simplifying the equation, we can solve for Ti:

m1 * c1 = (m2 * c2 + m3 * c3)

Ti = Tf - (m1 * c1) / (m2 * c2 + m3 * c3)

Now we can substitute the known values into the equation to find the initial temperature of the steel part.

Calculation

Let's calculate the initial temperature of the steel part using the given values and the formulas mentioned above.

m1 = 150 grams

c1 = 0.45 J/g°C

Tf = 60 degrees Celsius

m3 = 250 grams

c3 = 0.39 J/g°C

Substituting these values into the equation:

Ti = 60 - (150 * 0.45) / (m2 * 2.1 + 250 * 0.39)

We still need to determine the mass of the kerosene (m2) to calculate the initial temperature of the steel part.

Unfortunately, the given information does not provide the necessary data to determine the mass of the kerosene. Therefore, we cannot calculate the exact initial temperature of the steel part.

Conclusion

In conclusion, we are unable to determine the initial temperature of the steel part without knowing the mass of the kerosene. The given information is insufficient to solve the problem.