Question

The specific heat capacity of water is known to be 1 cal/(g⁰C). 

A system of units uses a unit of mass that equals 500g, a unit of length that equal 50cm and a unit of time that equal 100s. It also uses to Fahrenheit scale of temperature in which there are 180 divisons between the temperature of melting ice (0°C) and that of (normal) boiling point of water (= 100°C).

The value of the specific heat capacity of water, in this hypothetical system, would be:

(1) 9.3 x 10⁷ units of the hypothetical system

(2) 1.9 x 10⁷ units of the hypothetical system

(3) 1.9 x 10^-1 units of the hypothetical system

(4) 9.3 x 10^-1 units of the hypothetical system

Answer 1

(1) 9.3 x 10⁷ units of the hypothetical system

We have,

[Specific heat capacity] = Heat energy / Mass change in temperature

= [ML²T^-2/MK] = [L²T^-2K^-1]

∴ Unit of specific heat capacity in SI units / Unit of specific heat capacity in the hypothetical system

= (m/50cm)² (s/100s)^-2 (kelvin degree/Fahrenheit degree)^-1

Now 100 divisions on the celsius scale 

(= 100 divisions on the kelvin scale) 

= 180 divisions on the Fahrenheit scale 

∴ 1 division on the kelvin scale 

= 1.8 divisions on the Fahrenheit scale

∴ Above ratio = (100cm / 50cm)² (1/100)^-2 (1.8°F / 1°F)^-1

= 4 x 10⁴ x 1/1.8

= 2.22 x 10⁴

Now specific heat capacity of water = 1 cal / (g°C)

= 4.25 / 10^-3 kg(K) = 4.2 x 10³ J / kg (K) = 4.2 x 10³ SI Units

= 4.2 x 10³ x (2.22 x 10⁴) units in the hypothetical system

= 9.324 x 10⁷ units in the hypothetical system

= 9.3 x 10⁷ units