A cup of teat containing 400g of liquid at 20 degrees celsius was heated. If 58.5 kj of heat was absorbed by the tea, what would the final temperature be? Assume that the specific heat of the tea is the same as that of water (4.18).
I am not sure how to start this problem.
You know the specific heat formula of
q = mass x specific heat x (Tfinal-Tinitial). Start there. Substitute and solve for Tf
so I did 4.18 = 400 X 58.5 X (Tf-30)
4.18 =I amnot sure what to do on this side. Do I multiply 400, 58.5 and 30?
4.18 is the specific heat (and it states that in the problem), not q.
I did 400/58.5 (4.18) and got 28.6 degrees celsius. Was I suppose to use the 20 degrees that was stated in the problem?
To solve this problem, we can use the equation for heat transfer:
Q = mcΔT
Where:
Q is the heat absorbed or released by the substance (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
To start, let's convert the given 400g of liquid into kilograms. Since there are 1000g in a kilogram, 400g is equal to 0.4kg.
Now we can substitute the values into the equation:
58.5 kJ = (0.4 kg) * (4.18 J/g°C) * ΔT
First, let's convert 58.5 kJ into joules. Since there are 1000J in a kJ, 58.5 kJ is equal to 58,500 J.
58,500 J = (0.4 kg) * (4.18 J/g°C) * ΔT
Now, let's solve for ΔT:
ΔT = (58,500 J) / (0.4 kg * 4.18 J/g°C)
Calculating this, we get: ΔT ≈ 34.99°C
So, the final temperature of the tea would be 20°C + 34.99°C = approximately 54.99°C (rounded to two decimal places).
Therefore, the final temperature of the tea would be approximately 54.99°C.