Home
Class 11
PHYSICS
If an avarage person jogs, he produces 1...

If an avarage person jogs, he produces `14.5xx10^(4) cal//min`. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming `1 kg` requites `580xx10^(3)` cal for evaporation) is

A

`0.25 kg`

B

`2.25 kg`

C

`0.05 kg`

D

`0.20 kg`

Text Solution

Verified by Experts

The correct Answer is:
A

Amount of sweat evaporated/ minute
`=("sweat produced"//"minute")/("N o. of cals required for evaporation"//kg)=(14.5xx10^(4))/(580xx10^(3))=(145)/(580)= 0.25kg`
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    PRADEEP|Exercise Multiple choice questions.|96 Videos
  • THERMODYNAMICS

    PRADEEP|Exercise Interger Type questions|11 Videos
  • THERMODYNAMICS

    PRADEEP|Exercise Problems for practice|54 Videos
  • SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

    PRADEEP|Exercise Assertion- Reason Type questions|20 Videos
  • WORK, ENERGY AND POWER

    PRADEEP|Exercise Assertion-Reason Type Questions|24 Videos

Similar Questions

Explore conceptually related problems

If an average jogs, he produces 14.5xx10^(3) cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580xx10^(3) cal for evaporation) is

A metal rod AB of length 10x has its one end A in ice at 0^@C , and the other end B in water at 100^@C . If a point P one the rod is maintained at 400^@C , then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540cal//g and latent heat of melting of ice is 80cal//g . If the point P is at a distance of lambdax from the ice end A, find the value lambda . [Neglect any heat loss to the surrounding.]

A feverish person is running a temperature of 101^@F and is given paracetamol to lower his temperature. Due to medicine, there is an increase in rate of evaporation of sweat from his body. The mass of the person is 50 kg and the specific heat of human body is approximately about 1 cal g^(-1) ""^@C^(-1) . If the fever is brought down to 98^@F in 25 mins, the average rate of extra evaporation caused by the medicine is R in g "min"^(-1) , then the value of 87R/100 is (Latent heat of water at that temperature=580 g/cal)

A child running a temperature of 101^(@)F is given an antipyrine (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98^(@)F in 20 min , what is the average rate of extra evaporation caused by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. the mass of the child is 30 kg . the specific heat of human body is approximately the same as that of water, and latent heat of water at that temperature is about 580 cal//g .

A child running a temperature of 101^F is given and antipyrin (i.e. a madicine that lowers fever) which cause an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98^@F in 20 min., what is the averatge rate of extra evaporation caused, by the drug ? Assume the evaporation mechanism to the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal. g^(-1).

A thermally insulated vessel contains 150g of water at 0^(@)C . Then the air from the vessel is pumped out adiabatically. A fraction of water turms into ice and the rest evaporates at 0^(@)C itself. The mass of evaporated water will be closest to : (Latent heat of vaporization of water =2.10xx10^(6)jkg^(-1) and Latent heat of Fusion of water =3.36xx10^(5)jkg^(-1) )

The solar energy received by the Earth persquare metre per minute is 8.315 xx 10^(4)Jm^(-2)min^(-1) . If the radius of the Sun is 7.5 xx 10^(5) km and the distance of the Earth from the Sun is 1.5 xx10^(8) km, calculate the surface temperature of Sun. Assume the Sun as a perfect black body. Given that Stefen constanta sigma = 5.7 xx 10^(-8)Wm^(-2)K^(-4).

Assuming the radius of the earth to be 6.5 xx 10^(6) m. What is the time period T and speed of satellite for equatorial orbit at 1.4 xx 10^(3) km above the surface of the earth.