An 80.0-gram sample of a gas was heated from 25 degrees celsius to 225 degrees celsius. During this process, 346j of work was done by the system and its internal energy increased by 7785j. What is the specific heat of the gas?
delta E = delta H + work
7785 = q + (-346)
Solve for q.
Then q = mass x specific heat x (Tfinal-Tinitial)
You know q, mass, Tfinal and Tinitial. Solve for sp. h. .
To find the specific heat of the gas, we need to use the formula:
q = m * c * ΔT
Where:
- q is the heat absorbed or released by the gas
- m is the mass of the gas
- c is the specific heat capacity of the gas
- ΔT is the change in temperature
Given information:
- Mass of the gas, m = 80.0 grams
- Change in temperature, ΔT = (225°C - 25°C) = 200°C (convert to Kelvin by adding 273.15: 200 + 273.15 = 473.15K)
- Heat absorbed by the gas, q = increase in internal energy + work done by the system
= 7785J + 346J = 8131J
Now we can plug these values into the formula and solve for c:
8131J = 80.0 g * c * 473.15 K
Divide both sides by (80.0 g * 473.15 K):
c = 8131J / (80.0 g * 473.15 K)
Using a calculator, we can calculate the specific heat, c, of the gas.