A 0.0800-kilogram sample of a gas is heated from 25°C to 225°C. During this process, 340 J of work is done by the gas and its internal energy increases by 8450 J. How much heat is transferred to the gas?

What is the specific heat of the gas for this process?(J/(Kg.K))

dE - q _ w

w = -340 J
dE = 8450 J
Solve for q.

q = mass gas x specific heat x (Tfinal-Tinitial)
Substitute and solve for specific heat.

To determine the amount of heat transferred to the gas, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

Mathematically, this can be expressed as:

ΔU = Q - W,

where ΔU is the change in internal energy, Q is the heat added, and W is the work done.

In this case, we are given that the internal energy increases by 8450 J and the work done by the gas is 340 J.

Substituting these values into the equation, we have:

8450 J = Q - 340 J.

To solve for Q, we rearrange the equation:

Q = 8450 J + 340 J.

Adding the values, we find:

Q = 8790 J.

Therefore, the heat transferred to the gas is 8790 J.

To find the specific heat of the gas, we use the formula:

Q = mcΔT,

where Q is the heat transferred, m is the mass of the gas, c is the specific heat, and ΔT is the change in temperature.

Rearranging the equation to solve for c, we have:

c = Q / (m * ΔT).

We are given that the mass of the gas is 0.0800 kg, and the temperature change is from 25°C to 225°C.

Substituting these values into the equation, we have:

c = 8790 J / (0.0800 kg * (225°C - 25°C)).

Calculating the denominator first:

c = 8790 J / (0.0800 kg * 200°C).

Finally, evaluating the expression:

c = 8790 J / 16.0 kg°C.

Hence, the specific heat of the gas for this process is approximately 549.38 J/(kg·K).

To find the amount of heat transferred to the gas, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat transferred to the system plus the work done by the system. Mathematically, it is represented as:

ΔU = Q + W

Where:
ΔU is the change in internal energy of the gas,
Q is the heat transferred to the gas,
and W is the work done by the gas.

From the given information:
ΔU = 8450 J
W = 340 J

Substituting these values into the equation, we have:

8450 J = Q + 340 J

To find the value of Q, we can rearrange the equation as:

Q = 8450 J - 340 J
Q = 8110 J

Therefore, the amount of heat transferred to the gas is 8110 J.

Now, to find the specific heat of the gas for this process (c), we can use the formula:

Q = m * c * ΔT

Where:
Q is the heat transferred to the gas (8110 J),
m is the mass of the gas (0.0800 kg),
c is the specific heat of the gas,
and ΔT is the change in temperature (225°C - 25°C = 200°C).

Substituting these values into the equation, we have:

8110 J = 0.0800 kg * c * 200°C

To isolate c, we can rearrange the equation as:

c = 8110 J / (0.0800 kg * 200°C)
c = 50.69 J/(kg·K)

Therefore, the specific heat of the gas for this process is 50.69 J/(kg·K).