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Exercises · 1.3

Q.A calorie is a unit of heat (energy in transit) and it equals about 4.2 J4.2\ \text{J} where 1 J=1 kg m2 s−21\,\text{J} = 1\ \text{kg m}^2\,\text{s}^{-2}. Suppose we employ a system of units in which the unit of mass equals α\alpha kg, the unit of length equals β\beta m, the unit of time is γ\gamma s. Show that a calorie has a magnitude 4.2 α−1 β−2 γ24.2\ \alpha^{-1}\,\beta^{-2}\,\gamma^{2} in terms of the new units.

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The key idea is dimensional conversion: a calorie has dimensions [ML2T−2][M L^2 T^{-2}], so when base units change by factors α,β,γ\alpha, \beta, \gamma, the numerical value transforms by α−1β−2γ2\alpha^{-1} \beta^{-2} \gamma^{2}, giving 4.2 α−1β−2γ24.2\,\alpha^{-1}\beta^{-2}\gamma^{2} in the new system.

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 [ML2T−2][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 = α\alpha kg (so it's α\alpha times larger than the kg)
  • length unit = β\beta m (so it's β\beta times larger than the metre)
  • time unit = γ\gamma s (so it's γ\gamma times larger than the second)

Since energy has dimensions [ML2T−2][M L^2 T^{-2}], the numerical value in the new system = (old value) × (mass factor)−1^{-1} × (length factor)−2^{-2} × (time factor)+2^{+2}.

Step-by-step derivation

  1. Write the given conversion in SI units

    1 calorie=4.2 J1\ \text{calorie} = 4.2\ \text{J} and 1 J=1 kg m2s−21\ \text{J} = 1\ \text{kg m}^2 \text{s}^{-2}.

    So dimensionally, 1 calorie=4.2 [ML2T−2]1\ \text{calorie} = 4.2\ [M L^2 T^{-2}] in SI.

  2. Define the new units

    Let:

    • M′=α kgM' = \alpha\ \text{kg} (new unit of mass)
    • L′=β mL' = \beta\ \text{m} (new unit of length)
    • T′=γ sT' = \gamma\ \text{s} (new unit of time)

    This means:

    • 1 kg=1α M′1\ \text{kg} = \frac{1}{\alpha}\ M'
    • 1 m=1β L′1\ \text{m} = \frac{1}{\beta}\ L'
    • 1 s=1γ T′1\ \text{s} = \frac{1}{\gamma}\ T'

  3. Convert the calorie into new units

    Start from 1 cal=4.2 kg m2s−21\ \text{cal} = 4.2\ \text{kg m}^2 \text{s}^{-2}. Substitute the expressions above:

1 cal=4.2(1α M′)(1β L′)2(1γ T′)−21\ \text{cal} = 4.2 \left( \frac{1}{\alpha}\ M' \right) \left( \frac{1}{\beta}\ L' \right)^2 \left( \frac{1}{\gamma}\ T' \right)^{-2}

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

  1. Simplify the powers

=4.2⋅1α⋅1β2⋅γ2⋅M′L′2T′−2= 4.2 \cdot \frac{1}{\alpha} \cdot \frac{1}{\beta^2} \cdot \gamma^{2} \cdot M' L'^2 T'^{-2}

The combination M′L′2T′−2M' 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).

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

1 calorie=4.2 α−1β−2γ2 (new energy units)1\ \text{calorie} = 4.2\ \alpha^{-1} \beta^{-2} \gamma^{2}\ \text{(new energy units)}

Watch out

A common mistake is to get the sign of the exponent on γ\gamma wrong. Remember: time appears in the denominator (T−2T^{-2}), so when the unit gets larger by γ\gamma, the numerical factor must increase by γ2\gamma^2 — hence the positive exponent.

Tip

The pattern is simple: for a quantity with dimensions [MaLbTc][M^a L^b T^c], the conversion factor is α−aβ−bγ−c\alpha^{-a} \beta^{-b} \gamma^{-c}. Here a=1a=1, b=2b=2, c=−2c=-2, so it's α−1β−2γ2\alpha^{-1} \beta^{-2} \gamma^{2}.

✓Final answer

The magnitude of a calorie in the new units is 4.2 α−1β−2γ24.2\ \alpha^{-1} \beta^{-2} \gamma^{2}.

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