Question

1）On a certain dry sunny day, a swimming pool's temperature would rise by 1.40°C if not for evaporation. What fraction of the water must evaporate to carry away precisely enough heat to keep the temperature constant?

2）A piece of iron block moves across a rough horizontal surface before coming to rest. The mass of the block is 3.3 kg, and its initial speed is 3.2 m/s. How much does the block's temperature increase, if it absorbs 74% of its initial kinetic energy as internal energy? The specific heat of iron is 452 J/(kg · °C).

°C

Answer 1

(1)

Amount of heat energy required to raise the temperature of pool by 1.4 C

Q = Mtotal*Cw*T = Mtotal*4.187*1.4 ...............(1)

here,

Mtotal = mass of water in the pool

Cw = specific heat capacity of water = 4.187 KJ/Kg-K

Now,

Rather than increasing the temperature of water by 1.4 C, obtained heat energy is used to evaporate some amount of water having mass Mvap.

hence,

Qvap = Mvap*(hfg) = Mvap*2430 ......................(2)

here,

Mvap = mass of water vapor evaporated

(hfg) = Latent heat of vaporization = 2430 KJ/Kg

{from equation 1 & 2}

Mvap*2430 = Mtot*4.187*1.4

Mvap/Mtot = 2.412*10^-3 Ans

(2)

Kinetic energy of iron block = 1/2*(m)*(v^2) = 0.5*(3.3)*(3.2^2) = 16.896 Joule

as per question 74 % of K.E. is absorbed as internal enery.

hence

increase in internal energy of block = 0.74 * K.E. = 0.74*16.896 = 12.5 Joule

also,

increase in internal energy = M*C*T

12.5 = 3.2*452*T

T = 8.644*10^-3 C Ans

here,

T = increase in temperature of block

C = specific heat capacity of block

M = mass of blcok