In this post, we will solve a few numerical problems using the **First Law of Thermodynamics**.

## Numerical problems on first law of thermodynamics

**Question number 1)**

Complete the values in Table 1 (all values are in kJ).

**Solution:**

**As per the data given in Table 1: **** W = 6 kJ,**** U _{f} = 35 kJ, **

**ΔU = 24 kJ**

Q = ?

U_{i} = ?

**First Law of Thermodynamics** is expressed as: ** Δ**U = U

_{f }− U

_{i}= Q + W

**=> Q = ΔU – W = 24 – 6 = 18 kJ …… (1)**

Again, ** Δ**U = U

_{f }− U

_{i}

**=> U**

_{i}= U_{f}–*Δ*U = 35 – 24 = 11 kJ ………. (2)** Question number 2)**

A foam cup is filled with hot water and allowed to cool while being stirred by a paddlewheel.

Initially, the water has an internal energy of 200 kJ, and while cooling it loses 150 kJ of heat.

The paddlewheel does 25 kJ of work on the water.

a) Calculate the change in the internal energy of the water.

b) Calculate the final internal energy of the water.

**Solution:**

Known values:

Ui = 200 kJ

W = +25 kJ (positive as work is done on the water)

Q =– 150 kJ (negative as the energy is being lost or removed from the system)

a) Calculating ** Δ**U :

**U = Q + W**

*Δ*= (− 150 ) + (+ 25)

= − 125 kJ

**The change in the internal energy of the water is – 125 kJ. The negative sign means it has lost 125 kJ.**

b) Calculating U_{f} :

** Δ**U = U

_{f }− U

_{i}

U_{f} = ** Δ**U + U

_{i}= -125 + 200 = 75 kJ

**The final internal energy of the water is 75 kJ.**

**Question number 3)**

A pump full of compressed gas is allowed to expand and 80 kJ of work is done by the gas on an object in the lab. At the same time, the gas is warmed by the addition of 100 kJ of heat energy. If the initial internal energy of the gas is 500 kJ, calculate the final internal energy.

**Solution:**

Known values:

U_{i}= 500 kJ

W = – 80 kJ (negative as work is done by the gas)

Q = +100 kJ (positive as the energy is added to the system)

a) Calculate ** Δ**U :

**U = Q + W**

*Δ***U = + 100 + (− 80)**

*Δ***U = + 20 kJ**

*Δ*The positive sign means it has gained 20 kJ.

b) Calculate U

_{f}:

**U = U**

*Δ*_{f }− U

_{i}

=> U_{f} = ** Δ**U + U

_{i}

U

_{f}= +20 + 500

= + 520 kJ

The final internal energy is +520 kJ.