contains 200 grams of oil that's 30 degrees celsius
they add 100 grams of that oil that's 70 degrees celsius
the balance temperature is 40 degrees celsius
what's the specific heat capacity
of the oil
It can't happen if you are adding the same oil and the specific heat is constant from 30 to 100 C.
The 200 g of cold oil would gain 200*10*C calories and the 100 g of hot oil would lose 100*30C calories, where c is the specific heat.
They have to be equal.
To find the specific heat capacity of the oil, we can use the formula:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we have two different temperatures of oil that are being mixed. We can use the concept of heat transfer to find the specific heat capacity.
Let's start by calculating the heat transferred when adding 100 grams of oil at 70 degrees Celsius to 200 grams of oil at 30 degrees Celsius:
Q1 = mcΔT1
Q1 = (100g)(c)(40°C - 30°C)
Q1 = 100c(10°C)
Q1 = 1000c
Next, we need to calculate the heat transferred when the resulting oil mixture reaches 40 degrees Celsius:
Q2 = mcΔT2
Q2 = (300g)(c)(40°C - Tfinal)
Q2 = 12000c - 300cTfinal
Since the system reaches equilibrium, the heat transferred in Q1 should be equal to Q2:
Q1 = Q2
1000c = 12000c - 300cTfinal
Now, we can rearrange the equation to solve for the specific heat capacity (c):
1000c = 12000c - 300cTfinal
300cTfinal = 11000c
Tfinal = 11000/300
Tfinal = 36.67°C
Now that we know the equilibrium temperature (Tfinal), we can substitute it back into the equation to find the specific heat capacity (c):
1000c = 12000c - 300c(36.67°C)
1000c = 12000c - 11001c
1000c + 11001c = 12000c
12001c = 12000c
c = 12001c / 12001
c ≈ 1 J/g°C
Therefore, the specific heat capacity of the oil is approximately 1 J/g°C.