recall and use the first law of thermodynamics ∆U = q + W expressed in terms of the increase in internal energy, the heating of the system (energy transferred to the system by heating) and the work done on the system

First Law of Thermodynamics

The first law states that the total energy of an isolated system is constant. Energy can be transferred into or out of a system as heat or as work, but the change in the internal energy of the system equals the sum of these transfers.

Mathematical Form

\$\Delta U = q + W\$

where

• \$\Delta U\$ – change in internal energy of the system (J)
• \$q\$ – heat transferred to the system (J)
• \$W\$ – work done on the system (J)

Sign Conventions Used in A‑Level Physics

In the Cambridge A‑Level syllabus the following sign convention is adopted:

Components of the Equation

• Increase in internal energy (\$\Delta U\$) – the net result of all energy transfers.
• Heating of the system (\$q\$) – energy transferred by temperature difference.
• Work done on the system (\$W\$) – mechanical energy transferred by volume change, electrical work, etc.

Work Done on a Gas

For a quasi‑static process involving a gas, the work done on the system is

\$W = -\int{Vi}^{V_f} p\,\mathrm{d}V\$

The negative sign arises from the sign convention: when the gas expands (\$\mathrm{d}V>0\$) the system does work on the surroundings, so \$W\$ is negative.

Heat Transfer

When a substance of mass \$m\$ and specific heat capacity \$c\$ undergoes a temperature change \$\Delta T\$, the heat transferred is

\$q = mc\Delta T\$

For phase changes, \$q = mL\$, where \$L\$ is the latent heat.

Using the First Law – Step‑by‑Step Procedure

1. Identify the system and its surroundings.
2. Determine the direction of heat flow and assign the correct sign to \$q\$.
3. Determine the type of work involved (e.g., \$pV\$ work, electrical work) and calculate \$W\$ with the appropriate sign.
4. Substitute \$q\$ and \$W\$ into \$\Delta U = q + W\$ to find the change in internal energy.
5. If required, relate \$\Delta U\$ to other state variables (e.g., temperature) using \$ \Delta U = nC_V\Delta T\$ for an ideal gas.

Example Problem

A 0.50 kg block of aluminium (specific heat \$c = 900\;\text{J kg}^{-1}\text{K}^{-1}\$) is heated from \$20^\circ\text{C}\$ to \$80^\circ\text{C}\$ while being compressed by a piston so that \$20\;\text{J}\$ of work is done on it. Calculate the change in internal energy of the block.

Solution:

1. Heat transferred to the block:

   \$q = mc\Delta T = (0.50\;\text{kg})(900\;\text{J kg}^{-1}\text{K}^{-1})(80-20\;\text{K}) = 27\,000\;\text{J}\$

   Since heat flows into the system, \$q = +27\,000\;\text{J}\$.

2. Work done on the block is given as \$20\;\text{J}\$, so \$W = +20\;\text{J}\$.
3. Apply the first law:

   \$\Delta U = q + W = 27\,000\;\text{J} + 20\;\text{J} = 27\,020\;\text{J}\$

4. Thus the internal energy of the aluminium increases by \$2.70\times10^{4}\;\text{J}\$.

Common Pitfalls

• Mixing up the sign convention for work – remember that work done by the system is negative.
• Neglecting the work term when the problem involves volume change.
• Using \$q = mc\Delta T\$ for a phase change; the correct expression is \$q = mL\$.
• Assuming \$\Delta U = 0\$ for an isothermal process without checking whether the system is ideal and whether work and heat truly cancel.

Key Points to Remember

• The first law links heat, work, and internal energy: \$\Delta U = q + W\$.
• Positive \$q\$ = heat added to the system; positive \$W\$ = work done on the system.
• For \$pV\$ work, \$W = -\int p\,\mathrm{d}V\$; expansion gives negative work, compression gives positive work.
• Always state the system clearly before assigning signs.
• Use appropriate formulas for \$q\$ (sensible heating, latent heat) and \$W\$ (mechanical, electrical, surface work) before substituting into the first‑law equation.