How to calculate dtheta/dt for newtons law of cooling in practicals?
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Newton's Law of cooling :
According to Newton's law of cooling, The rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surrounding.
dQ/dt= K (T2-T1)-----------------(1)
Procedure :
The calorimeter is filled with two third of liquid and heated up to temperature of 80 °C .
This liquid will acts as hot body which will be cooling.Thermometer is inserted in the calorimeter.
When the temperature reaches to 70°C stop watch will be started to note down time
.Time should be noted for every 5°C fall of temperature upto room temperature.
Observation :
Readings are tabulated :
******************************************
S.no Time Temperature
(sec) °c
****************************************
1.
2.
3.
4.
5.
*****************************************
Draw the graph between temperature on Y axis and time along X axis
.calculate the slope to tangents drawn at various temperature
.Slope dQ/dt=------------
According to Newton's law of cooling, The rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surrounding.
dQ/dt= K (T2-T1)-----------------(1)
Procedure :
The calorimeter is filled with two third of liquid and heated up to temperature of 80 °C .
This liquid will acts as hot body which will be cooling.Thermometer is inserted in the calorimeter.
When the temperature reaches to 70°C stop watch will be started to note down time
.Time should be noted for every 5°C fall of temperature upto room temperature.
Observation :
Readings are tabulated :
******************************************
S.no Time Temperature
(sec) °c
****************************************
1.
2.
3.
4.
5.
*****************************************
Draw the graph between temperature on Y axis and time along X axis
.calculate the slope to tangents drawn at various temperature
.Slope dQ/dt=------------
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