28 chapter homework physics solution

28 chapter homework physics solution

These solutions for Heat Transfer are extremely popular among Class 11 Science students for Physics Heat Transfer Solutions come handy for quickly completing your homework and preparing for exams. Why don't we write d Q d t? This is because the amount of heat crossing through any cross section is a function of many variables like temperature difference, area of cross-section, etc. So, we cannot write it as a complete derivative with respect to time. If yes, why does its temperature not fall further?

Physics chapter 28 homework solution for resume book kellogg

These solutions for Heat Transfer are extremely popular among Class 11 Science students for Physics Heat Transfer Solutions come handy for quickly completing your homework and preparing for exams. Why don't we write d Q d t? This is because the amount of heat crossing through any cross section is a function of many variables like temperature difference, area of cross-section, etc.

So, we cannot write it as a complete derivative with respect to time. If yes, why does its temperature not fall further?

Yes, the body will radiate. However, its temperature will not fall down with time because as the temperature of the surroundings is greater than the temperature of the body so, its rate of absorption will be greater than its rate of emission.

Why does blowing over a spoonful of hot tea cools it? Does evaporation play a role? Does radiation play a role? Here, major role is played by convection. When we blow air over a spoonful of hot tea, the air coming from our mouth has less temperature than the air above the tea. Since hot air has less density, it rises up and cool air goes down. In this way, the tea cools down. We know that any hot body radiates.

So, the spoonful of tea will also radiate and as the temperature of the surrounding is less then the tea, the tea will cool down with time. Evaporation is also involved in this. On blowing over the hot tea, rate of evaporation increases and the cools down. On a hot summer day we want to cool our room by opening the refrigerator door and closing all the windows and doors.

Will the process work? When the door of the refrigerator is left open in a closed room, the heat given out by the refrigerator to the room will be more than that taken from the room. Therefore, instead of decreasing, the temperature of the room will increase at a slower rate. On a cold winter night you are asked to sit on a chair. Would you like to choose a metal chair or a wooden chair? Both are kept in the same lawn and are at the same temperature.

We will prefer to seat on a wooden chair because as the conductivity of wood is poorer than that of metal, heat flow from our body to the chair will be less in case of a wooden chair. Will the temperatures equalise by radiation? Yes, the temperature of the balls can be equalised by radiation. This is because both the spheres will emit radiations in all the directions at different rates.

The ball kept at the temperature of K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of K. Also, it losses energy by radiation. Similarly, the ball kept at the temperature of K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of K. A time comes when the temperature of both the bodies becomes equal. An ordinary electric fan does not cool the air, still it gives comfort in summer because it circulates the air present in the room.

Due to this, evaporation takes place and we feel cooler. Yet an animal there would freeze to death and not boil. So, very less molecules of air collide with the body of the animal and transfer very less amount of heat.

That is why the animal present there would freeze to death instead boiling. Standing in the sun is more pleasant on a cold winter day than standing in shade.

Is the temperature of air in the sun considerably higher than that of the air in shade? The heat coming from the sun to us is through the radiation. On colder winter days, when we stand in shade, we do not get the heat of the sun from the radiation. Though we feel cool in the shade, the temperature of the air in shady as well as non-shady regions is the same. During night, the earth's surface radiates infrared radiation of larger wavelength.

Gas molecules in the air absorb some of this energy and radiate energy of their own in all directions. Also, water molecules, like the vapour that makes the clouds, absorb more frequencies of infrared energy than clear air does. Both these factors contribute to the fact that clouds radiate more heat in all directions including the earth than clear air does.

In turn, this makes the overall temperature on the earth warmer when there is a cloud cover. The heat energy radiated by the earth is reflected back to earth. Due to this, cloudy nights are warmer than the nights with clean sky. A white colour dress reflects almost all the radiations falling on it. So, it does not absorb any heat from the sunlight and we feel more comfortable in it. On the other hand, a dark colour dress absorbs maximum radiation falling on it. So, we feel hot in a dark coloured dress during summers.

The thermal conductivity of a rod depends on a length b mass c area of cross section d material of the rod. For example, metals are much better conductors than non-metals because metals have large number of free electron that can move freely anywhere in the body of the metal and carry thermal energy from one place to other.

Also, 2 copper rods having different lengths and areas of cross-section have same thermal conductivity that depends only on the number of free electrons in copper. In a room containing air, heat can go from one place to another a by conduction only b by convection only c by radiation only d by all the three modes.

In this process, the average position of a molecule does not change. Hence, there is no mass movement of matter. In convection, heat is transferred from one place to other by actual motion of particles of the medium. When water is heated, hot water moves upwards and cool water moves downwards. In radiation process, transfer of heat does not require any material medium. For a room containing air, heat can be transferred via radiation no medium required and convection by the movement of air molecules and by conduction due to collision of hot air molecules with other molecules.

Since the temperature of the solid is less than the surroundings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The thermal radiation emitted by a body is proportional to T n where T is its absolute temperature. The value of n is exactly 4 for a a blackbody b all bodies c bodies painted black only d polished bodies only.

This law holds true for all the bodies. The thermal radiation emitted in a given time by A and B are in the ratio a 1 : 1. One end of a metal rod is kept in a furnace. In steady state, the temperature of the rod a increases b decreases c remain constant d is nonuniform.

Newton's law of cooling is a special case of a Wien's displacement law b Kirchhoff's law c Stefan's law d Planck's law. Suppose a body at temperature T is kept in a room at temperature T 0. A hot liquid is kept in a big room. Its temperature is plotted as a function of time. Which of the following curves may represent the plot?

The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T 4 , where T is the initial temperature of the liquid. As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease.

Therefore, the plot of temperature with time is best represented by the curve in option a. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly a a straight line b a circular are c a parabola d an ellipse. As the temperature decreases, the rate of loss will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid.

So, in steady state, the difference between the temperatures of the two will become zero. A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The logarithm converts the fourth power dependence into a linear dependence with some coefficient property of log.

So, the plot satisfying all the above properties will be a straight line. One end of a metal rod is dipped in boiling water and the other is dipped in melting ice. So, with time, the temperature of the rod will increase from the end dipped in boiling water to the end dipped in melting, until it comes in equilibrium with its surroundings.

In steady state, the temperature of the rod is nonuniform and constant, maximum at the end dipped in boiling water and minimum at the end dipped in melting ice. Equilibrium means that the system is stable.

So, all the macroscopic variables describing the system will not change with time. Hence, the temperature of the rod will become constant once equilibrium is reached, but its value is different at different positions of the rod. A blackbody does not a emit radiation b absorb radiation c reflect radiation d refract radiation. A black body is the one that absorbs all the radiation incident on it. So, a black body does not reflect and refract radiation.

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Mastering Physics Solutions. Solution: When two waves interfere destructively at one place, then at some other place, these waves interfere constructively. The energy at the point of destructive interference at one place is always balanced by that at constructive interference. In destructive interference, the net energy of the resultant wave is less than the sum of energies of two individual waves, which interfere destructively to give destructive interference.

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HC Verma Class 12 Physics Part-2 Solutions for Chapter 28 - Heat Transfer

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