Contents
Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What is Water Equivalent and SI Unit?
Definition: Thermal capacity of a body is defined as the quantity of heat required to raise its temperature by unity.
Let the mass of a body be m and its specific heat be s. Heat gained by the body to raise its temperature by unity, is H = m ᐧ s ᐧ 1 = ms. By definition, this is the thermal capacity. It is usually denoted by the symbol C.
Hence, thermal capacity of a body = mass of the body × specific heat capacity of its material.
Units of thermal capacity:
Relation between thermal capacity and specific heat: We know, thermal capacity of a body
= mass of the body × specific heat of its material or, specific heat of the material
\(=\frac{\text { thermal capacity of the body }}{\text { mass of the body }}\)
If the mass of the body is 1 unit, specific heat of the material = thermal capacity of the body.
Hence, thermal capacity of unit mass of a body is the specific heat of its material.
Water Equivalent
Definition: Water equivalent of a given body is the mass of water for which the rise in temperature is the same as that for the body, when the amount of heat supplied is the same for both.
Let, the mass of a body be ni and Ìts specific heat be s.
Hence, the amount of heat required to raise the temperature of the body by t,
H = ms
If this amount of heat is also required for a mass W of water
(specific heat = sw) for the same rise in temperature then
H = mst = W × sw × t
∴ W = \(\frac{m s}{s_w}\)
So, water equivalent of the body, W = \(\frac{m s}{s_w}\)
In CGS system and SI, the value of sw are 1 cal ᐧ g-1 ᐧ °C-1 and 4186 J ᐧ kg-1 ᐧ K-1 respectively. So the expressions of W in CGS and SI are,
W = ms ………. (2)
and W = \(\frac{m s}{4186}\) ………. (3)
Hence, heat gained or lost by a body,
H = Wswt …….. (4)
So, heat gained or lost by a body = water equivalent of the body × specific heat of water × rise or fall in temperature.
‘Water equivalent of a body is 10 g ’ means, heat required to raise the temperature of the body by 1 ° C can raise the temperature of 10 g of water by 1° C.
Numerical Examples
Example 1.
Specific heat capacity of aluminium is 0.21 cal ᐧ g-1 ᐧ °C-1. What will be the thermal capacity and water equivalent of an aluminium strip of mass 200 g?
Solution:
Here, m = 200 g and s = 0.21 cal ᐧ g-1 ᐧ °C-1
Thermal capacity of the aluminium strip = m ᐧ s = 200 × 0.21 = 42 cal ᐧ C-1.
Water equivalent of the aluminium strip
= \(\frac{m \cdot s}{s_w}\) = 200 × 0.21 [∵ sw = 1 cal ᐧ g-1 ᐧ °C-1]
= 42g
Example 2.
The ratio of the densities of the materials of two bodies is 2 : 3 and that of their specific heat capacities is 0.12 : 0.09. Find the ratio of their thermal capacities per unit volume.
Solution:
Let the densities of two substances be ρ1 and ρ2 and their specific heats be s1 and s2, respectively.
According to the question, ρ1 – ρ2 = 2 : 3 and s1 : s2 = 0.12 : 0.09 .
Let their thermal capacities per unit volume be H1 and H2 respectively. As mass per unit volume is density ρ, we have thermal capacity per unit volume = ρ ᐧ s.
∴ \(\frac{H_1}{H_2}\) = \(\frac{\rho_1 s_1}{\rho_2 s_2}\) = \(\frac{2}{3}\) × \(\frac{0.12}{0.09}\) = \(\frac{8}{9}\)
Example 3.
Thermal capacities of mercury and glass of the same volume are equal. Densities of mercury and glass are 13.6 g ᐧ cm-3 and 2.5 g ᐧ cm-3 respectively. If the specific heat capacity of mercury is 0. 03 cal ᐧ g-1 ᐧ °C-1, find that of glass.
Solution:
Let the volume of each of mercury and glass be V and specific heat of glass be s.
Thermal capacity of mercury = V × 13.6 × 0.03 and thermal capacity of glass = T × 2.5 × s
According to the problem,
V × 13.6 × 0.03 = V × 2.5 × s
∴ s = \(\frac{13.6 \times 0.03}{2.5}\) = 0.163 g-1 ᐧ °C-1
Example 4.
600 g of water at 30°C is kept in a vessel of water equivalent 60 g. If the vessel is supplied heat at the rate of 100 cal ᐧ s-1, how much time will the water take to reach its boiling point?
Solution:
Heat absorbed by the water
= 600 × 1 × (100 – 30) = 600 × 70 = 42000 cal
Heat absorbed by the container
= 60 × 1 × (100 – 30) = 60 × 70 = 4200 cal
∴ Total heat absorbed = 42000 + 4200 = 46200 cal
∴ Required time = \(\frac{46200}{100}\) = 462 s = 7 min 42 s.
Example 5.
Specific gravities of two liquids are 0.8 and 0.5. The thermal capacity of 3 L of the first one is equal to that of 2 L of the second one. Compare their specific heats. [HS(XI) ’07]
Solution:
Volume of the first liquid = 3 L = 3000 cm3
∴ Mass of the first liquid, m1 = 3000 × 0.8 g
Volume of the second liquid = 2 L = 2000 cm3
∴ Mass of the second liquid, m2 = 2000 × 0.5 g
As the thermal capacities of the two liquids are equal,
m1s1 = m1s2
or, \(\frac{s_1}{s_2}\) = \(\frac{m_1}{m_2}\) = \(\frac{2000 \times 0.5}{3000 \times 0.8}\) = \(\frac{2 \times 5}{3 \times 8}\) = \(\frac{5}{12}\)