An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 8305 J. What is the specific heat of the gas?
Scroll down and find the post I made for Sam. Same problem. It isn't the first post you see for Sam the the one for same below that.
To find the specific heat of the gas, we can make use of the first law of thermodynamics, which states that the change in internal energy (∆U) of a system is equal to the heat transferred (Q) to the system minus the work (W) done by the system.
Mathematically, this can be represented as:
∆U = Q - W
In this case, we are given:
∆U = 8305 J (increase in internal energy)
W = 346 J (work done by the system)
We need to find Q (heat transferred to the system) in order to calculate the specific heat of the gas.
Using the first law of thermodynamics equation, we rearrange it to solve for Q:
Q = ∆U + W
Q = 8305 J + 346 J
Q = 8651 J
Now that we have the value of Q, we can calculate the specific heat (C) of the gas using the formula:
Q = m * C * ∆T
Where:
m = mass of the gas
C = specific heat of the gas
∆T = change in temperature
We are given:
m = 80.0 g (mass of the gas)
∆T = 225 °C - 25 °C = 200 °C
Plugging in the values:
8651 J = 80.0 g * C * 200 °C
To solve for C, divide both sides of the equation by (80.0 g * 200 °C):
C = 8651 J / (80.0 g * 200 °C)
C = 0.541 J/(g °C) (rounded to three significant figures)
So, the specific heat of the gas is approximately 0.541 J/(g °C).