Advertisements
Advertisements
प्रश्न
63.2 g of copper at 50°C can just melt 3.8g of ice. If the specific latent heat of ice is 336 J/g, find the specific heat capacity of copper.
Advertisements
उत्तर
Let the specific heat capacity of copper be s
∴ Heat lost by copper = 63.2 × s × (50 - 0) = 3160 s joule
Heat gained by ice = 3.8 × 336 joule
Since Heat gained = Heat lost
We have 3160s = 3.8 × 336 = 1276.8
Or s = (1276.8/3160) K/g °C = 0.404 J/g °C.
APPEARS IN
संबंधित प्रश्न
Heat energy is supplied at a constant rate to 100g of ice at 0 °C. The ice is converted into water at 0° C in 2 minutes. How much time will be required to raise the temperature of water from 0 °C to 20 °C? [Given: sp. heat capacity of water = 4.2 J g-1 °C-1, sp. latent heat of ice = 336 J g-1].
The specific heat capacity of water is :
What is the specific heat capacity of boiling water?
Name the substance which has maximum specific heat capacity.
Write an expression for the heat energy liberated by a hot body.
How much heat energy is released when 5 g of water at 20° C changes to ice at 0° C?
[Specific heat capacity of water = 4.2 J g-1 ° C-1 Specific latent heat of fusion of ice = 336 J g-1]
From the options given below the specific heat of _______ is maximum.
The cold object the hot object enclosed in one box of heat-resistant material.
- What changes will occur in the two objects when temperature flows from those objects?
- Which principle can show that the energy exchange takes place between two objects only when kept in isolated system?
Numerical Problem.
How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C? (Specific latent heat of fusion of water = 3,34,000 J/kg, Specific heat capacity of water = 4200 JKg–1K–1).
