Q.A calorie is a unit of heat (energy in transit) and it equals about $4.2\ \text{J}$ where $1\,\text{J} = 1\ \text{kg m}^2\,\text{s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha$ kg, the unit of length equals $\beta$ m, the unit of time is $\gamma$ s. Show that a calorie has a magnitude $4.2\ \alpha^{-1}\,\beta^{-2}\,\gamma^{2}$ in terms of the new units.

Concept: Unit Conversion — When you change the base units, the numerical value of a physical quantity changes inversely with the size of each unit raised to its dimension.

Reasoning:

1. The dimension of energy (and heat) is $[M L^2 T^{-2}]$. In SI, 1 calorie = $4.2\ \text{kg m}^2 \text{s}^{-2}$.

2. In the new system, 1 new unit of mass = $\alpha$ kg, so 1 kg = $\alpha^{-1}$ new mass units.

   Similarly, 1 m = $\beta^{-1}$ new length units, and 1 s = $\gamma^{-1}$ new time units.

3. Substitute into the SI expression:

Why this works: the logic of unit conversion

Every physical quantity has dimensions — a combination of mass, length, and time. A calorie is a unit of energy, and energy has dimensions $[M L^2 T^{-2}]$. When we change the base units, the numerical value of a fixed physical quantity changes inversely to the size of the units.

Think of it this way: if you measure a table's length in metres and get 2, then switch to centimetres (which are 100 times smaller), the number becomes 200 — larger because the unit is smaller. The conversion factor is the reciprocal of the unit-size factor.

Here, the new units are:

• mass unit = $\alpha$ kg (so it's $\alpha$ times larger than the kg)
• length unit = $\beta$ m (so it's $\beta$ times larger than the metre)
• time unit = $\gamma$ s (so it's $\gamma$ times larger than the second)

Since energy has dimensions $[M L^2 T^{-2}]$, the numerical value in the new system = (old value) × (mass factor)$^{-1}$ × (length factor)$^{-2}$ × (time factor)$^{+2}$.

Step-by-step derivation

1. Write the given conversion in SI units

   $1\ \text{calorie} = 4.2\ \text{J}$ and $1\ \text{J} = 1\ \text{kg m}^2 \text{s}^{-2}$.

   So dimensionally, $1\ \text{calorie} = 4.2\ [M L^2 T^{-2}]$ in SI.

2. Define the new units

   Let:

   • $M' = \alpha\ \text{kg}$ (new unit of mass)
   • $L' = \beta\ \text{m}$ (new unit of length)
   • $T' = \gamma\ \text{s}$ (new unit of time)

   This means:

   • $1\ \text{kg} = \frac{1}{\alpha}\ M'$
   • $1\ \text{m} = \frac{1}{\beta}\ L'$
   • $1\ \text{s} = \frac{1}{\gamma}\ T'$

3. Convert the calorie into new units

   Start from $1\ \text{cal} = 4.2\ \text{kg m}^2 \text{s}^{-2}$. Substitute the expressions above:

Notice the time term: $\text{s}^{-2}$ means we take the reciprocal of the square of the conversion.

4. Simplify the powers

The combination $M' L'^2 T'^{-2}$ is exactly 1 unit of energy in the new system (by definition, since it has the same dimensions as a joule in the new units).

5. Read off the numerical value Therefore, in the new system:

Method: Dimensional Analysis for Unit Conversion

This method uses the fact that physical quantities have dimensions that remain invariant under a change of units. We express the given quantity in terms of base dimensions, then convert each base unit to the new system.

Steps

Step 1: Write the dimension of energy (calorie or joule)

From $1\ \text{J} = 1\ \text{kg m}^2\ \text{s}^{-2}$, the dimension of energy is:

Step 2: Express the conversion factor for each base unit

• New unit of mass $= \alpha\ \text{kg}$

  $\Rightarrow 1\ \text{kg} = \alpha^{-1}$ (new mass units)

• New unit of length $= \beta\ \text{m}$

  $\Rightarrow 1\ \text{m} = \beta^{-1}$ (new length units)

• New unit of time $= \gamma\ \text{s}$

  $\Rightarrow 1\ \text{s} = \gamma^{-1}$ (new time units)

Step 3: Substitute into the dimensional formula

Since $1\ \text{calorie} = 4.2\ \text{J}$, and $1\ \text{J} = 1\ \text{kg m}^2\ \text{s}^{-2}$, we replace each base unit:

Step 4: Write the final result

Key Insight

The numerical value of a physical quantity changes inversely with the size of the unit. Since the new mass unit is $\alpha$ times larger than kg, the numerical value in new units becomes $\alpha^{-1}$ times the old value — and similarly for length and time, following the dimensional exponents.

Common Mistakes & How to Avoid Them

1. Confusing the direction of conversion (inverse vs. direct)

The Mistake:

Students often think: "Since 1 new unit of mass = $\alpha$ kg, then 1 kg = $\alpha$ new units." This is wrong.

Why it's wrong:

If the new unit is larger (e.g., $\alpha > 1$), then the number of new units in 1 kg should be smaller.

• Example: If $\alpha = 2$ (1 new mass unit = 2 kg), then 1 kg = $0.5$ new units = $\alpha^{-1}$ new units.

How to avoid:

Always ask: "Is the new unit bigger or smaller than the old unit?"

• 1 new unit = $\alpha$ old units → 1 old unit = $\frac{1}{\alpha}$ new units = $\alpha^{-1}$ new units.

Correct relation:

• 1 kg = $\alpha^{-1}$ new mass units
• 1 m = $\beta^{-1}$ new length units
• 1 s = $\gamma^{-1}$ new time units

2. Forgetting to convert all three base units

The Mistake:

Students convert mass correctly but forget to convert length and time in the expression $1\ \text{J} = 1\ \text{kg m}^2 \text{s}^{-2}$.

Why it's wrong:

The joule involves three base units — missing even one gives the wrong exponent.

How to avoid:

Write the dimensional formula explicitly:

Then convert each dimension separately:

So:

3. Getting the sign of the time exponent wrong

The Mistake:

Students write $\gamma^{-2}$ instead of $\gamma^{2}$.

Why it's wrong:

Time appears in the denominator ($s^{-2}$). When converting, the factor for $s^{-2}$ becomes $(\gamma^{-1})^{-2} = \gamma^{2}$.

How to avoid:

Treat the exponent carefully:

• $s^{-2}$ means $(\text{time})^{-2}$
• 1 s = $\gamma^{-1}$ new time units
• So $s^{-2} = (\gamma^{-1})^{-2} = \gamma^{2}$

Quick check: If $\gamma > 1$ (new time unit is longer), the numerical value of energy in new units should be larger — which $\gamma^{2}$ gives.

4. Forgetting to multiply by the numerical factor (4.2)

The Mistake:

Students show the conversion factor correctly but forget that 1 calorie = $4.2$ J, so the final answer must include $4.2$.

How to avoid:

Always start with:

Then convert the joule. Never drop the $4.2$.

Final correct expression:

Quick Summary Checklist

Final answer to verify against: