A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.

Open in App

Solution

Now, $Δ Q = 2625 cal$

$Δ U = 5000 J$

From graph,

$Δ W = 200 \times 10^{3} \times 0.03$

$= 6000 J$

Now, $Δ Q = Δ W + Δ U$

$\Rightarrow 2625 cal. = 6000 J + 5000 J$

$\Rightarrow J = \frac{11000}{2625} = 4.19 J / cal$

Suggest Corrections

Similar questions

Q. A gas is taken along the path AB as shown in figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.

Q. The P - V diagram of a system undergoing a thermodynamic process is shown in figure. Work done by the system in going from

A \to B \to C

is 30 J and 40 J heat is given to the system. The change in internal energy of the gas if the gas is directly taken from A to C is

When a system is taken through the process abc shown in figure, 80 J of heat is absorbed by the system and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during the process ?

A gas is taken through a cyclic process ABCA as shown in figure. If 2.4 cal of heat is given in the process, what is the value of J ?

Q. If change in internal energy is given by

- 120 J

. The work done by the system is 280 J. Calculate the heat associated with the process.

Join BYJU'S Learning Program

A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.

Now, ΔQ=2625 cal ΔU=5000 J From graph, ΔW=200×103×0.03 =6000 J Now, ΔQ=ΔW+ΔU ⇒2625 cal.=6000 J+5000 J ⇒J=110002625=4.19 J/cal

Now, $Δ Q = 2625 cal$

$Δ U = 5000 J$

From graph,

$Δ W = 200 \times 10^{3} \times 0.03$

$= 6000 J$

Now, $Δ Q = Δ W + Δ U$

$\Rightarrow 2625 cal. = 6000 J + 5000 J$

$\Rightarrow J = \frac{11000}{2625} = 4.19 J / cal$