100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in

Q: Why do we ignore the expansion of water?

In reality, water expands slightly when heated from 30°C to 50°C. However, this expansion is so minuscule that the work done ($p\Delta V$) against atmospheric pressure is negligible compared to the heat added. Ignoring it simplifies the calculation without significantly affecting the result.

Q: Why is the temperature difference in Celsius used directly in the formula which has Kelvin?

A temperature difference ($\Delta T$) of 1°C is exactly equal to a temperature difference of 1 K. Therefore, $50^\circ\text{C} - 30^\circ\text{C} = 20^\circ\text{C}$, which is exactly $20\text{ K}$.

Q: What if the expansion was not ignored?

If expansion wasn't ignored, you would need the coefficient of volume expansion for water to calculate $\Delta V$. Then, you'd calculate the work done $W = p\Delta V$ and subtract it from the total heat supplied to find the exact change in internal energy.

Q: Why did we convert mass to kilograms?

The specific heat capacity is given in J/kg/K. To maintain dimensional consistency and obtain the energy in Joules, the mass must be in kilograms.

Q: Is the heat supplied equal to the change in internal energy for all liquids?

Yes, for most liquids and solids under normal conditions, the volume change during heating is negligible. Thus, the work done is almost zero, making the heat supplied practically equal to the change in internal energy.

Question

100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K)

Options:

(A) 8.4 kJ

(B) 84 kJ

(C) 2.1 kJ

(D) 4.2 kJ

Sigma Percentile · Accepted Answer

### The Setup: Heating Water

Imagine a beaker containing $100	ext{ g}$ of water. We are heating it, and its temperature is rising from $30^\circ	ext{C}$ to $50^\circ	ext{C}$. This is a classic thermodynamics scenario where heat is being transferred to a system, causing its internal state to change.

### The Hidden Clue: Ignoring Expansion

Look closely at the question... there is a very important phrase: **'ignoring the slight expansion'**. This implies that the volume of the water remains constant throughout the heating process.

So, the change in volume, $\Delta V$, is exactly zero.

Now, what is the formula for work done in thermodynamics? It is given by $W = p\Delta V$. Since $\Delta V = 0$, the work done by the system against the surroundings is also zero ($W = 0$).

### The First Law of Thermodynamics

Let's recall the First Law of Thermodynamics, which is essentially the law of conservation of energy for thermal systems. It states that the heat supplied to a system is used to do work and to increase its internal energy:

$$\Delta Q = \Delta U + W$$

Because the work done is zero, all the heat supplied goes entirely into increasing the internal energy. So, we can simplify our master equation to:

$$\Delta U = \Delta Q$$

### Calculating the Internal Energy

And what is the formula to calculate the heat supplied, $\Delta Q$? It is given by the specific heat formula:

$$\Delta Q = mC\Delta T$$

Here, $m$ is the mass, $C$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Let's carefully substitute the values. The mass is $100	ext{ g}$, which must be converted to standard SI units, giving $0.1	ext{ kg}$. The specific heat is given as $4184	ext{ J/kg/K}$. And the temperature change is $50 - 30 = 20	ext{ K}$.

$$\Delta U = (0.1	ext{ kg}) 	imes (4184	ext{ J/kg/K}) 	imes (20	ext{ K})$$

Don't make a silly mistake in the calculation. $0.1$ multiplied by $20$ gives $2$. And twice of $4184$ is $8368	ext{ Joules}$.

$$\Delta U = 8368	ext{ J}$$

The options are provided in kilo Joules. So, converting $8368	ext{ Joules}$ gives $8.368	ext{ kJ}$. This is approximately **$8.4	ext{ kJ}$**. It's a beautifully simple application of the First Law!

Animated Solution for Physics - Thermodynamics: 100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K)

Select Answer:

Visualized Solution

System Setup

Volume Constraint

Work Done

First Law of Thermodynamics

Heat Supplied

Substitution

Calculation

Final Answer

The Sigma Insight: First Law of Thermodynamics

The Setup

The Hidden Clue

The First Law of Thermodynamics

Calculating the Internal Energy

Similar Questions

The internal energy change (in J) when $90 g$ of water undergoes complete evaporation at $10 0^{\circ} C$ is ........., (Given : $Δ H_{vap}$ for water at $373 K = 41 kJ/mol$ , $R = 8.314 J K^{- 1} mol^{- 1}$ )

Assuming that water vapour is an ideal gas, the internal energy change ( $Δ E$ ) when 1 mole of water is vaporised at 1 bar pressure and $10 0^{\circ} C$ , (Given : molar enthalpy of vaporisation of water at 1 bar and 373 K = $41 kJ mol^{- 1}$ and $R = 8.3 J mol^{- 1} K^{- 1}$ ) will be

A system does $200 J$ of work and at the same time absorbs $150 J$ of heat. The magnitude of the change in internal energy is ......... $J$ . (Nearest integer)

For an ideal gas

For water $Δ_{vap} H = 41 kJ mol^{- 1}$ at $373 K$ and $1 bar$ pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ......... $kJ mol^{- 1}$ . [Use $R = 8.3 J mol^{- 1} K^{- 1}$ ]

The internal energy change when a system goes from state $A$ to $B$ is $40 kJ/mol$ . If the system goes from $A$ to $B$ by a reversible path and returns to state $A$ by an irreversible path, what would be the net change in internal energy?

At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is ............

Following figure shows two processes A and B for a gas. If $Δ Q_{A}$ and $Δ Q_{B}$ are the amount of heat absorbed by the system in two cases, and $Δ U_{A}$ and $Δ U_{B}$ are changes in internal energies respectively, then

List-I describes thermodynamic processes in four different systems. List-II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List-I

List-II