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Question: If 500 calories of heat energy is added to a system and the system does 350 calories of work on the ...

If 500 calories of heat energy is added to a system and the system does 350 calories of work on the surroundings, what is the energy change of the system?

Explanation

Solution

This question is based on the mathematical form of the first law of thermodynamics which states that energy can neither be created nor be destroyed but can be transformed from one form to another.

Complete answer:
Let us consider the initial energy of the given system to be U1{U_1} . Say the system is supplied with some amount of heat. Let this heat added be q. Now the energy of the system becomes U2{U_2} . So we can write an expression as
U2=U1+q.................................Equation(1){U_2} = {U_1} + q.................................Equation (1)
Suppose some work w is being done on this system. Then, the RHS of equation (1) becomes
U2=U1+q+w..........................Equation(2){U_2} = {U_1} + q + w.......................... Equation (2)
Using the above equation, we can obtain the expression for energy change of the system as:
U2U1=q+w{U_2} - {U_1} = q + w
ΔU=q+w\Delta U = q + w\xrightarrow{{}} Equation (3) is the mathematical expression of the first law of thermodynamics.
Given, 500 calories of heat is added to the system. As heat is being added to the system, q=+500q = + 500 . From the question, we know that the system does 350 calories of work on the surroundings. As work is being done by the system, w=350w = - 350 . Substituting these values in equation (3) we get
ΔU=+500+(350)\Delta U = + 500 + ( - 350)
ΔU=+500350=150\Delta U = + 500 - 350 = 150 Calories

Note: The sign conventions to be used while solving numericals based on the first law of thermodynamics are as follows:
If heat is absorbed by the system, q is positive and if heat is evolved by the system q is negative. For work done on the system w is positive and if work is done by the system then w is negative.