Case study (4 Marks)

Question 14 Marks

A person's skin is more severely burnt when put in contact with 1g of steam at 100°C than when put in contact with 1g of water at 100°C. Explain.

Answer

The skin felts more burning as latent heat is involved in steam at 100°c and thus total heat of steam is greater than water at 100°c.

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Question 24 Marks

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10kg of water and 0.2g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decrease by 5°C. Specific heat capacity of water $= {4200J kg^{-1}}^\circ C^{-1}$ and latent heat of vaporization of water $= 2.27 \times 10^6\ J\ kg^{-1}.$

Answer

Given,
Specific heat of water, ${S = 4200\ J\ kg^{-1}}^\circ C^{-1}$
Latent heat of vapourisation of water, $L = 2.27 \times 10^6\ J\ kg^{-1}$
Mass, $M = 0.2g = 0.0002kg$
Let us first calculate the amount of energy required to decrease the temperature of 10kg of water by 5°C.
$U_1 = 10 \times 4200J/kg^\circ C \times 5^\circ C$
$U_1 = 210,000 = 21 \times 10^4 J$
Let the time in which the temperature is decreased by 5°C be t.
Energy required per second for evaporation of water (at the rate of 0.2g/sec) is given by
$U_2 = ML$
$U_2 = (2 \times 10^{-4} ) \times (2.27 \times 10^6) = 454J$
Total energy required to decrease the temperature of the water $= 454 × t$
$= 21 \times 10^4 J$
Now,
$\text{t}=\frac{21\times10^4}{454}$ seconds
The time taken in minutes is given by,
$\text{t}=\frac{21\times10^4}{454\times60}=7.7\text{ minute/s}$
$\therefore$ The time required to decrease the temperature by 5°C is 7.7 minutes.

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Physics Case study (4 Marks)

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