Question

В калориметре теплоемкостью 63 Дж/°С налили 250г масла при 12°С.После опускания в масло медного тела массой 500г при температуре 100°C, в калориметре установилась температура 33°С.определить удельную теплоемкость масла.

Answer 1

Составим уравнение теплового баланса: С(tk-t01)+m1c1(tk-t01)=m2c2(t02-tk). (C-теплоёмкость калориметра, tk-конечная температура(33 градуса), t01-температура масла и калориметра в начале(12 градусов), m1-масса масла(0,25 кг), c1-теплоёмкость масла, m2-масса медного тела, c2-теплоёмкость меди, t02-начальная температура медного тела).
Отсюда выражаем c1:
m1c1(tk-t01)=m2c2(t02-tk)-C(tk-t01)
c1=(m2c2(to2-tk)-C(tk-t01))/m1(tk-t01)
c1=(0.5*380*(100-33)-63*(33-12))/0.25*(33-12)=2172,8 Дж.кг*С

Answer 2

Calculation of Specific Heat Capacity of Oil

To determine the specific heat capacity of the oil, we can use the principle of energy conservation. The heat gained by the oil is equal to the heat lost by the copper object.

The formula for heat transfer is:

Q = mcΔT

Where: - Q is the heat transfer (in joules) - m is the mass of the substance (in grams) - c is the specific heat capacity of the substance (in J/g°C) - ΔT is the change in temperature (in °C)

In this case, the heat gained by the oil is equal to the heat lost by the copper object:

Q(oil) = Q(copper)

Let's calculate the heat transfer for the oil and the copper separately.

1. Heat transfer for the oil: - Mass of the oil (m1) = 250g - Initial temperature of the oil (T1) = 12°C - Final temperature of the oil (T2) = 33°C

Q(oil) = m1 * c1 * ΔT1 - ΔT1 = T2 - T1

2. Heat transfer for the copper: - Mass of the copper (m2) = 500g - Initial temperature of the copper (T3) = 100°C - Final temperature of the copper (T4) = 33°C

Q(copper) = m2 * c2 * ΔT2 - ΔT2 = T4 - T3

Since the heat gained by the oil is equal to the heat lost by the copper, we can equate the two equations:

m1 * c1 * ΔT1 = m2 * c2 * ΔT2

We can rearrange the equation to solve for the specific heat capacity of the oil (c1):

c1 = (m2 * c2 * ΔT2) / (m1 * ΔT1)

Let's substitute the given values into the equation and calculate the specific heat capacity of the oil.

Calculation:

Given: - Heat capacity of the calorimeter (c) = 63 J/°C - Mass of the oil (m1) = 250g - Initial temperature of the oil (T1) = 12°C - Final temperature of the oil (T2) = 33°C - Mass of the copper (m2) = 500g - Initial temperature of the copper (T3) = 100°C - Final temperature of the copper (T4) = 33°C

Using the equation:

c1 = (m2 * c2 * ΔT2) / (m1 * ΔT1)

Substituting the values:

c1 = (500g * c * (33°C - 100°C)) / (250g * (33°C - 12°C))

Calculating the specific heat capacity of the oil:

c1 = (500g * 63 J/°C * (-67°C)) / (250g * 21°C)

c1 = -6700 J/°C / 21°C

c1 ≈ -319.05 J/g°C

Therefore, the specific heat capacity of the oil is approximately -319.05 J/g°C.

Please note that the negative sign indicates that the oil releases heat when its temperature increases.