Advertisements
Advertisements
Question
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
Advertisements
Solution
Heat energy lost by the vessel and water contained in it in cooling the water from 50°C to 5°C is used in heating ice to melt it and then to raise its temperature from 0°C to 5°C.
Now, heat energy lost by the copper vessel is
Qc = mcccΔt = 100 x 0.4 x (50 - 5)
Qc = 1800 J
Similarly, heat energy lost by water is
Qw = mwcwΔt = 150 x 4.2 x (50 - 5)
Qw = 28350 J
Hence, the total heat energy lost is
QL = 1800 + 28350 = 30150 J
Let m g of ice be used to cool water. So, heat gained by ice is
QI = mLice + mcwΔt
QI = 336m + m x 4.2 x 5 = 336m + 21m = 357mJ
Therefore, from the principle of calorimetry, the mass of ice is
QL = QI
∴ 357m = 30150
∴ m = `30150/357` = 84.45 g
APPEARS IN
RELATED QUESTIONS
A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2 of other substance of specific heat capacity c2 at a lower temperature t2. Deduce the expression for the temperature t of the mixture. State the assumption made, if any.
A solid of mass 80 g at 80°C is dropped in 400 g water at 10°C. If final temp. is 30°C, find the sp. heat cap. of the solid.
Write the approximate values of the specific latent heat of vaporization of steam.
Explain, why temperature in hot summer, falls sharply after a sharp shower?
Calculate the ratio of two specific heats of polyatomic gas molecules.
Read this activity and answer the following questions.
- Take three spheres of iron, copper and lead. the lead of equal mass.
- Put all the three spheres in boiling water in the beaker for some time.
- Take the three spheres out of the water.
- All the spheres will be at a temperature 100 °C.
- Put them immediately on the thick slab of wax.
- Note, the depth that each of the sphere goes into the wax.
Questions:
- Which property is determined from this activity?
- Give name to that property.
- Explain the term principal of heat exchange with the help of this activity.
An office room contains about 4000 moles of air. The change in the internal energy of this much air when it is cooled from 34° C to 19° C at a constant pressure of 1.0 atm is (Use `gamma_"air"` = 1.4 and Universal gas constant = 8.314 J / mol K) ____________.
Prove the Mayer's relation `C_p - C _v = R/J`
