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A body of mass 25kg is dragged on a rough horizontal road for one hour with a speed of 20km/hr.
If the coefficient of friction is 0.5 and half of the heat produced is absorbed by the body, the rise in its temperature is (specific heat of body = 0.1cal/g ºC, g = 10m/s2)
Select one:
If the coefficient of friction is 0.5 and half of the heat produced is absorbed by the body, the rise in its temperature is (specific heat of body = 0.1cal/g ºC, g = 10m/s2)
Select one:
- a)59.5ºC
- b)84.5ºC
- c)39ºC
- d)119ºC
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A body of mass 25kg is dragged on a rough horizontal road for one hour...
Distance traveled in one hour,
S = 20km
Now,⇒
The correct answer is: 119ºC
S = 20km
Now,⇒
The correct answer is: 119ºC
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Community Answer
A body of mass 25kg is dragged on a rough horizontal road for one hour...
To find the rise in temperature, we need to calculate the heat produced and then divide it by the mass of the body.
First, we need to find the force of friction acting on the body. The formula for frictional force is given by:
Frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the body, which is given by:
Weight = mass * acceleration due to gravity
Weight = 25kg * 9.8m/s^2 = 245N
Frictional force = 0.5 * 245N = 122.5N
The work done by friction can be calculated using the formula:
Work done = force * distance
The distance traveled in one hour at a speed of 20 km/hr is:
Distance = speed * time = 20 km/hr * 1 hour = 20 km
Converting the distance to meters:
Distance = 20 km * 1000 m/km = 20,000 m
Work done = 122.5N * 20,000m = 2,450,000 J
Since half of the heat produced is absorbed by the body, the heat produced is:
Heat produced = 2,450,000 J * 2 = 4,900,000 J
To convert the heat produced from Joules to calories, we divide by the conversion factor:
Heat produced = 4,900,000 J / 4.18 J/cal = 1,172,248 cal
Finally, to find the rise in temperature, we divide the heat produced by the mass of the body:
Rise in temperature = Heat produced / (mass * specific heat)
The mass of the body is 25 kg, and the specific heat is 0.1 cal/g. Converting the mass to grams:
Mass = 25 kg * 1000 g/kg = 25,000 g
Rise in temperature = 1,172,248 cal / (25,000 g * 0.1 cal/g) = 4.69°C
Therefore, the rise in temperature of the body is 4.69°C.
First, we need to find the force of friction acting on the body. The formula for frictional force is given by:
Frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the body, which is given by:
Weight = mass * acceleration due to gravity
Weight = 25kg * 9.8m/s^2 = 245N
Frictional force = 0.5 * 245N = 122.5N
The work done by friction can be calculated using the formula:
Work done = force * distance
The distance traveled in one hour at a speed of 20 km/hr is:
Distance = speed * time = 20 km/hr * 1 hour = 20 km
Converting the distance to meters:
Distance = 20 km * 1000 m/km = 20,000 m
Work done = 122.5N * 20,000m = 2,450,000 J
Since half of the heat produced is absorbed by the body, the heat produced is:
Heat produced = 2,450,000 J * 2 = 4,900,000 J
To convert the heat produced from Joules to calories, we divide by the conversion factor:
Heat produced = 4,900,000 J / 4.18 J/cal = 1,172,248 cal
Finally, to find the rise in temperature, we divide the heat produced by the mass of the body:
Rise in temperature = Heat produced / (mass * specific heat)
The mass of the body is 25 kg, and the specific heat is 0.1 cal/g. Converting the mass to grams:
Mass = 25 kg * 1000 g/kg = 25,000 g
Rise in temperature = 1,172,248 cal / (25,000 g * 0.1 cal/g) = 4.69°C
Therefore, the rise in temperature of the body is 4.69°C.
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A body of mass 25kg is dragged on a rough horizontal road for one hour with a speed of 20km/hr.If the coefficient of friction is 0.5 and half of the heat produced is absorbed by the body, the rise in its temperature is (specific heat of body = 0.1cal/g ºC, g = 10m/s2)Select one:a)59.5ºCb)84.5ºCc)39ºCd)119ºCCorrect answer is option 'D'. Can you explain this answer?
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