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प्रश्न
Ice cream appears colder to the mouth than water at 0℃. Give reason.
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उत्तर
This is because 1 g of ice at 0°C takes 336 J of heat energy from the mouth to melt at 0°C. Thus, mouth loses an additional 336 J of heat energy for 1 g of ice at 0°C than for 1g of water at 0°C. Therefore, cooling produced by 1 g of ice at 0°C is more than for 1g of water at 0°C.
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संबंधित प्रश्न
A copper vessel of mass 100 g contains 150 g of water at 50°C. How much ice is needed to cool it to 5°C?
Given: Specific heat capacity of copper = 0.4 Jg-1 °C-1
The Specific heat capacity of water = 4.2 Jg-1 °C-1
The Specific latent heat of fusion ice = 336 Jg-1
What is meant by specific heat capacity?
A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg °C. It contains 250 gm of liquid at 30°C having specific heat of 0.4 kcal/kg °C. If we drop a piece of ice of mass 10 g at 0°C, What will be the temperature of the mixture?
Study the following procedure and answer the questions below:
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note, the depth that each sphere goes into the wax.
i) Which property of substance can be studied with this procedure?
ii) Describe that property in minimum words.
iii) Explain the rule of heat exchange with this property.
(i) State whether the specific heat capacity of a substance remains the same when its state changes from solid to liquid.
(ii) Give one example to support your answer.
A piece of iron of mass 2.0 kg has a thermal capacity of 966 J/°C. What is its specific heat capacity in S.I. units?
Express the change in internal energy in terms of molar specific heat capacity.
We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron `(β_(vbrass) = (6 xx 10^(–5))/K and β_(viron) = (3.55 xx 10^(–5))/K)` to create a volume of 100 cc. How do you think you can achieve this.
