Calorimetry Water Equivalent
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 m and its specific heat be s.
Hence, the amount of heat required to raise the temperature of the body by t, H = mst
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 x sw x t
∴ W = \(\frac{m s}{s_w}\)…(1)
So, water equivalent of the body, W = \(\frac{m s}{s_w}\)
Water Equivalent Formula In Calorimetry Class 11 HBSE
In CGS system and SI, the value of sw are 1 cal · g-1 · °C-1 and 4186 J · kg-1 · 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 x specific heat of water x rise or fall in temperature.
‘Water equivalent of a body is 10 g ’ means, heat required to raise the temperature of the body by 10 C can raise the temperature of 10 g of water by 1° C.
Calorimetry Water Equivalent Numerical Examples
Example 1. Specific heat capacity of aluminium is 0.21 cal · g-1 · °C-1. What will be the thermal capacity and water equivalent of an aluminum strip of mass
200 g?
Solution:
Given
Specific heat capacity of aluminium is 0.21 cal · g-1 · °C-1.
Here, m = 200 g and s = 0.21 cal · g-1 · °C-1
Thermal capacity of the aluminum strip = m · s = 200 x 0.21 = 42 cal · °C-1.
Water equivalent of the aluminium strip
= \(\frac{m s}{s_w}\) = 200×0.21 [sw = 1 cal · g-1 · °C-1 ] = 42g
Water Equivalent In Calorimetry Class 11 HBSE Notes
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:
Given
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.
Let the densities of two substances be ρ1 and ρ2, and their specific heats be s1 and s2, respectively.
According to the 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 · s.
∴ \(\frac{H_1}{H_2}=\frac{\rho_1 s_1}{\rho_2 s_2}=\frac{2}{3} \times \frac{0.12}{0.09}=\frac{8}{9} .\)
Derivation Of Water Equivalent In Calorimetry Class 11
Example 3. The thermal capacities of mercury and glass of the same volume are equal. Densities of mercury and glass are 13.6 g ·cm-3 and 2.5 g · cm-3 respectively. If the specific heat capacity of mercury is 0.03 cal · g-1 · °C-1, find that of glass.
Solution:
Given
The thermal capacities of mercury and glass of the same volume are equal. Densities of mercury and glass are 13.6 g ·cm-3 and 2.5 g · cm-3 respectively. If the specific heat capacity of mercury is 0.03 cal · g-1 · °C-1
Let the volume of each of mercury and glass be V and specific heat of glass be s.
Thermal capacity of mercury = Vx 13.6 x 0.03 and thermal capacity of glass =V x 2.5 x s
According to the problem, Vx 13.6 x 0.03 = V x 2.5 x s
∴ s = \(\frac{13.6 \times 0.03}{2.5}=0.163 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \text {. }\)
Water Equivalent And Specific Heat Class 11 Physics HBSE
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 · s-1, how much time will the water take to reach its boiling point?
Solution:
Given
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 · s-1
Heat absorbed by the water = 600 x 1 x (100 – 30) = 600 x 70 = 42000 cal
Heat absorbed by the container = 60 x 1 x (100-30) = 60 x 70 = 4200 cal
∴ Total heat absorbed = 42000 + 4200 = 46200 cal
∴ Required time = ^99^ = 462 s = 7 min 42 s.
Numerical Problems On Water Equivalent Class 11 HBSE
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 j that of 2 L of the second one. Compare their specific heats.
Solution:
Given
Specific gravities of two liquids are 0.8 and 0.5. The thermal capacity of 3 L of the first one is equal to j that of 2 L of the second one
Volume of the first liquid = 3 L = 3000 cm³
∴ Mass of the first liquid, m1 = 3000 x 0.8 g
Volume of the second liquid = 2 L = 2000 cm³
∴ Mass of the second liquid, m2 = 2000 x 0.5 g
As the thermal capacities of the two liquids arc equal, m1s1 = m2s2
or, \(\frac{s_1}{s_2}=\frac{m_2}{m_1}=\frac{2000 \times 0.5}{3000 \times 0.8}=\frac{2 \times 5}{3 \times 8}=\frac{5}{12} .\)