if 100 grams of water is at 100 degree celcius are poured into 100 grams turpentine at 9 degree celcius and the resulting temperature found to be 70 degree celcius. what is the specific heat of the turpentine?
To find the specific heat of the turpentine, we can use the principle of energy conservation, assuming no heat loss to the surroundings. The formula we can use is:
Q1 + Q2 = 0
where Q1 is the heat lost by the water, and Q2 is the heat gained by the turpentine.
To calculate Q1, we can use the specific heat capacity formula:
Q1 = mcΔT1
where m is the mass of the water (100 grams), c is the specific heat capacity of water (4.18 J/g°C), and ΔT1 is the change in temperature of the water (final temperature - initial temperature).
Q1 = 100g * 4.18 J/g°C * (70°C - 100°C) = -12520 J
Since Q1 is negative, it means the water loses heat.
To calculate Q2, we can use the same formula:
Q2 = mcΔT2
where m is the mass of the turpentine (100 grams), c is the specific heat capacity of the turpentine, and ΔT2 is the change in temperature of the turpentine (final temperature - initial temperature).
Q2 = 100g * c * (70°C - 9°C)
Now, since Q1 + Q2 = 0, we can substitute the values and calculate the specific heat capacity of the turpentine.
-12520 J + 100g * c * (70°C - 9°C) = 0
Simplifying the equation, we have:
100g * c * 61°C = 12520 J
Dividing both sides of the equation by 61°C * 100g, we find:
c ≈ 20.49 J/g°C
Therefore, the specific heat capacity of the turpentine is approximately 20.49 J/g°C.