Advertisements
Advertisements
प्रश्न
The temperature of 170 g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.
Given: Specific heat capacity of water = 4200 J kg-1 °C-1 and specific latent heat of ice = 336000 J kg-1.
Advertisements
उत्तर
Given mass of water = 170 g = 1.17 kg,
Initial temperature = 50°C
Fall in temperature = Δt = (50 - 5) = 45°C = 45K
Heat lost by water = mc Δt
= 0.17 × 4200 × 45
= 3.213 × 104 J
If m' kg ice is added, heat gained by it to melt to 0°C = m'L
= m'C Δt
= m' × 4200 × 5
= m' × 2.1 × 104 J
Total heat gained by ice
= 3.36 × 105 m' + 2.1 × 104 m'
= 3.57 × 105 m' J
By the principle of method of mixtures heat lost by water = heat gained by ice
⇒ 3.213 × 104 = 3.57 × 105 m'
⇒ m' = `(3.213 xx 10^4)/(3.57 xx 10^5)`
⇒ m' = 0.09 kg (90 g)
APPEARS IN
संबंधित प्रश्न
Write the approximate value of specific heat capacity of water in S.I. unit.
Write the expression for the heat energy Q received by the substance when m kg of substance of specific heat capacity c Jkg-1 k-1 is heated through Δt° C.
Explain the term boiling point ?
What is meant by global warming?
Define heat capacity.
A certain amount of heat Q will warm 1 g of material X by 3°C and 1 g of material Y by 4°C. Which material has a higher specific heat capacity?
Solve the following problem.
What is the specific heat of metal if 50 cal of heat is needed to raise 6 kg of the metal from 20°C to 62 °C?
State factors on which the amount of heat radiated by a body depends.
J/Kg °C is the unit of specific heat capacity.
A block of ice of mass 120 g at temperature 0°C is put in 300 gm of water at 25°C. The xg of ice melts as the temperature of the water reaches 0°C. The value of x is ______.
[Use: Specific heat capacity of water = 4200 Jkg-1K-1, Latent heat of ice = 3.5 × 105 Jkg-1]
