Effect of Heat CBSE Class 9 Physics Worksheets Set 05

Questions with Answers

Question 1. When a glass vessel is heated, its volume increases a little. By adding some mercury to the vessel, the increase in the volume of the vessel can be neutralised by the expansion of the mercury, so that the space above the mercury has a constant volume. Calculate the volume of mercury to be added as a fraction of the volume of the vessel (cubical expansivity of mercury 18 x 10⁻⁵ /°C, linear expansivity of glass 9 x 10⁻⁶ /°C).
Answer:
Cubical expansivity of glass = 3 x 9 x 10⁻⁶ = 27 x 10⁻⁶ /°C. Let the vessel volume be V and the mercury volume v, with a rise of temperature t.
Expansion of mercury = \( v \times 18 \times 10^{-5} t \); expansion of the vessel = \( V \times 27 \times 10^{-6} t \). Equating them gives \( \dfrac{v}{V} = \dfrac{27 \times 10^{-6}}{18 \times 10^{-5}} = 0.15 \).
The volume of mercury must be 3/20 (about 1/7) of the volume of the vessel.
In simple words: Mercury must expand as much as the vessel's capacity does, so that the empty space above it stays constant.

Teacher's Note:
1. The vessel's capacity expands as if it were solid glass.
2. The empty space above the mercury then keeps a constant volume at all temperatures.

Question 2. A 500 g solid is heated by a 30 W electric heater. The temperature of the solid rises at the rate of 5 K per minute. Calculate the heat capacity and the specific heat capacity of the solid.
Answer:
Heat supplied per minute = 30 x 60 = 1800 J, and this raises the temperature by 5 K.
Heat capacity = \( \dfrac{1800}{5} = 360 \) J/K.
Specific heat capacity = \( \dfrac{360}{0.5} = 720 \) J/kg K. Heat capacity 360 J/K; specific heat 720 J/kg K.
In simple words: Energy supplied per minute divided by the temperature rise gives the heat capacity.

Teacher's Note:
1. Power 30 W means 30 J every second.
2. Specific heat is the heat capacity per unit mass.

Question 3. A steel ball is immersed in alcohol. The apparent weight of the ball is W₁ at 0°C and W₂ at 59°C. Is W₂ greater than, equal to or less than W₁? (The coefficient of volume expansion of alcohol is greater than that of steel.)
Answer:
Apparent weight = actual weight - buoyant force, and the buoyant force = (volume of ball) x (density of alcohol) x g.
Between 0°C and 59°C alcohol expands much more than steel, so the density of alcohol decreases by a larger fraction than the volume of the ball increases. The buoyant force therefore decreases.
W₂ is greater than W₁.
In simple words: Hot alcohol is lighter, so it pushes up the ball less and the ball appears heavier.

Teacher's Note:
1. Buoyancy depends on the liquid's density.
2. The actual weight of the ball does not change.