Test: Specific Heat Capacity - EmSAT Achieve MCQ

• When water is sprinkled over a large area, the evaporation of the water takes place.
• When the water evaporates, it absorbs the heat from the surroundings because the latent heat of vaporization is large for water.
• Due to this absorption of energy, cooling takes place in the surroundings.

• The Ball was initially at a height 10 meters above the ground and when it bounced from the ground its height was 5.5 meters above the ground.
• So the energy absorbed and raised temperature of ball was the difference between these two potential energies of the ball which was actually a difference between the initial and final height of the ball.

Given Data:
h_initial = initial height = 10 m , m = 0.5 kg
⟹  h_final = final height = 5.5 m ,
C = 450 J/kg° C
m = 0.5 kg
∴ Q = mg (h_initial - h_final)
∴ mC⋅(ΔT) = mg (hinitial - hfinal)
∴ 0.5 × 450 × ΔT = 0.5 × 9.81 × (10 - 5.5)
∴ ΔT = 0.196 ° C  ≈ 0.2 ° C
⟹ The nearest value in options are 0.1 ° C so we have to choose it.

• As it is heat required for unit mass of the body so specific heat doesn’t depend on mass of that body.
• Specific heat capacity of anybody/material is a property of that body/material. It is same for each and every molecules of that body. So it doesn’t depend on the shape of the body. Thus  independent of mass and shape of the body.
• As per the definition of the specific heat, it is the heat required to raise the temperature by 1°C. Hence the specific heat doesn’t depend on the temperature of the body.

For diatomic molecules

Heat energy given by the metal piece is 750 kJ

750 = m c ΔT
= 4 × c × (115−15)
750 = 4 × c × 100

∴ c = 1.875 kJ/kg °C

The correct answer is an option (1).

Compressibility is the reciprocal of the bulk modulus of elasticity.
Compressibility (p) = 1/K, and K = bulk modulus of Elasticity

For isothermal process:

Differentiating equation (ii),
PdV + Vdp = 0
⇒ PdV = -Vdp

From equation (i) & (iii), we have
K = P

The energy produced by the heater is used to raise the temperature of liquid from 25°C to 31°C in 2 minutes.

Let heat capacity is H

Temperature change =  Δ T  = 31 ° C - 25 ° C = 6 ° C = 6 K

Heat required to raise temperature E  = H × 6 K -- (1)

Power = 1000 W

time t = 2 min = 120 s

Eneegy supplied by heater E = P × t = 1000 W × 120 s  -- (2)

Equating (1) and (2)

H × 6 K = 1000 W × 120 s 

So, the correct option is 2 × 10⁴ J/°C

Given - The mass of substance = m, temperature change = ΔT amount of heat absorbed or rejected by a substance = ΔQ 

• The amount of heat absorbed or rejected by a substance of mass 'm' and specific heat capacity 's', when it undergoes a temperature change 'ΔT' is equal to

⇒ Q = ms∆T
So specific heat capacity is,

Given that,
Mass of wood, m = 1 kg
Heat absorbed by wood for increasing its temperature by  25°C to 150°C, ΔQ = 200 kJ = 200000 J 
Change in temperature, ΔT = 150 - 25 = 125 °C 
Hence, from the above explanation, we can see that specific heat of any material can be calculated as 

• As it is heat required for unit mass of the body so specific heat doesn’t depend on the mass of that body.
• Specific heat capacity of anybody/material is a property of that body/material. It is same for each and every molecule of that body.
• So it doesn’t depend on the shape of the body. Thus independent of mass and shape of the body.
• As per the definition of the specific heat, it is the heat required to raise the temperature by 1°C. Hence the specific heat doesn’t depend on the temperature of the body.