We all know that hot water or milk when left on a table begins to cool gradually. Ultimately it attains the temperature of the surroundings. To study how a given body can cool on exchanging heat with its surroundings, let us perform the following activity.
Take some water, say 300 ml, in a calorimeter with a stirrer and cover it with two holed lid. Fix a thermometer through a hole in the lid and make sure that the bulb of thermometer is immersed in the water.
`"Note"` the reading of the thermometer. This reading `T_1` is the temperature of the surroundings. Heat the water kept in the calorimeter till it attains a temperature, say, `40 °C` above room temperature (i.e., temperature of the surroundings). Then stop heating the water by removing the heat source.
Start the stopwatch and note the reading of the thermometer after fixed interval of time, say after every one minute of stirring gently with the stirrer. Continue to note the temperature `(T_2)` of water till it attains a temperature about `5 °C` above that of the surroundings.
Then plot a graph by taking each value of temperature `color{blue}(ΔT = T_2 – T_1)` along y axis and the coresponding value of t along x-axis (Fig. 11.18).
From the graph you will infer how the cooling of hot water depends on the difference of its temperature from that of the surroundings.
You will also notice that initially the rate of cooling is higher and decreases as the temperature of the body falls.
The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature.
According to Newton’s law of cooling, the rate of loss of heat, `– dQ//dt` of the body is directly proportional to the difference of temperature `ΔT = (T_2–T_1)` of the body and the surroundings.
The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
`color{green}(- (dQ)/(dt) = k (T_2\ \T_1))`....(11.15)
where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass m and specific heat capacity s is at temperature `T_2.` Let `T_1` be the temperature of the surroundings. If the temperature falls by a small amount d `T_2` in time dt, then the amount of heat lost is
`color{navy}(dQ = ms dT_2)`
:. Rate of loss of heat is given by
`color{orange}((dQ)/(dt) =ms (dT_2)/(dt))`........(11.16)
From Eqs. (11.15) and (11.16) we have
`color{green}(-ms(dT_2)/(dT) = k (T_2-T_1))`
`color{orange}((dT_2)/(T_2_T_1)=-k/ms dt=-Kdt)`....(11.17)
where `K = k//m s`
On integrating, `color{red}(loge (T_2 – T_1) = – K t + c)`........(11.18)
or `color{orange}(T_2 = T_1 + C′ e^(–Kt); \ \ \ \ where C′ = e^c)`..... (11.19)
Equation (11.19) enables you to calculate the time of cooling of a body through a particular range of temperature.
For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.
Newton’s law of cooling can be verified with the help of the experimental set-up shown in Fig. 11.19(a). The set-up consists of a double walled vessel (V) containing water in between the two walls.
A copper calorimeter (C) containing hot water is placed inside the double walled vessel. Two thermometers through the corks are used to note the temperatures `T_2` of water in calorimeter and `T_1` of hot water in between the double walls respectively.
Temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between loge `(T_2–T_1)` and time (t). The nature of the graph is observed to be a straight line having a negative slope as shown in Fig. 11.19(b). This is in support of Eq. (11.18).
We all know that hot water or milk when left on a table begins to cool gradually. Ultimately it attains the temperature of the surroundings. To study how a given body can cool on exchanging heat with its surroundings, let us perform the following activity.
Take some water, say 300 ml, in a calorimeter with a stirrer and cover it with two holed lid. Fix a thermometer through a hole in the lid and make sure that the bulb of thermometer is immersed in the water.
`"Note"` the reading of the thermometer. This reading `T_1` is the temperature of the surroundings. Heat the water kept in the calorimeter till it attains a temperature, say, `40 °C` above room temperature (i.e., temperature of the surroundings). Then stop heating the water by removing the heat source.
Start the stopwatch and note the reading of the thermometer after fixed interval of time, say after every one minute of stirring gently with the stirrer. Continue to note the temperature `(T_2)` of water till it attains a temperature about `5 °C` above that of the surroundings.
Then plot a graph by taking each value of temperature `color{blue}(ΔT = T_2 – T_1)` along y axis and the coresponding value of t along x-axis (Fig. 11.18).
From the graph you will infer how the cooling of hot water depends on the difference of its temperature from that of the surroundings.
You will also notice that initially the rate of cooling is higher and decreases as the temperature of the body falls.
The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature.
According to Newton’s law of cooling, the rate of loss of heat, `– dQ//dt` of the body is directly proportional to the difference of temperature `ΔT = (T_2–T_1)` of the body and the surroundings.
The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
`color{green}(- (dQ)/(dt) = k (T_2\ \T_1))`....(11.15)
where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass m and specific heat capacity s is at temperature `T_2.` Let `T_1` be the temperature of the surroundings. If the temperature falls by a small amount d `T_2` in time dt, then the amount of heat lost is
`color{navy}(dQ = ms dT_2)`
:. Rate of loss of heat is given by
`color{orange}((dQ)/(dt) =ms (dT_2)/(dt))`........(11.16)
From Eqs. (11.15) and (11.16) we have
`color{green}(-ms(dT_2)/(dT) = k (T_2-T_1))`
`color{orange}((dT_2)/(T_2_T_1)=-k/ms dt=-Kdt)`....(11.17)
where `K = k//m s`
On integrating, `color{red}(loge (T_2 – T_1) = – K t + c)`........(11.18)
or `color{orange}(T_2 = T_1 + C′ e^(–Kt); \ \ \ \ where C′ = e^c)`..... (11.19)
Equation (11.19) enables you to calculate the time of cooling of a body through a particular range of temperature.
For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.
Newton’s law of cooling can be verified with the help of the experimental set-up shown in Fig. 11.19(a). The set-up consists of a double walled vessel (V) containing water in between the two walls.
A copper calorimeter (C) containing hot water is placed inside the double walled vessel. Two thermometers through the corks are used to note the temperatures `T_2` of water in calorimeter and `T_1` of hot water in between the double walls respectively.
Temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between loge `(T_2–T_1)` and time (t). The nature of the graph is observed to be a straight line having a negative slope as shown in Fig. 11.19(b). This is in support of Eq. (11.18).