NEWTON'S LAW OF COOLING Objective To study the relationship between the temperature of a hot body and its time of cooling by plotting a cooling curve. NAVIGATION PATH: http://amrita.olabs.edu.in/?sub=1&brch=5&sim=21&cnt=4 Theory This Law of Cooling is named after the famous English Physicist Sir Isaac Newton, who conducted the first experiments on the nature of cooling. Statement of the Law : According to Newton's Law of Cooling, the rate of cooling of a body is directly proportional to the difference in temperatures of the body (T) and the surrounding (To), provided difference in temperature should not exceed by 30"C. From the above statement, do (T - To) dt For a body of mass m, specific heat s, and temperature T kept in surrounding of temperature To; Q = msT Now, the rate of cooling, dQ dT de = ms- dt Hence, dT dt a (T - To) Since the mass and the specific heat of the body are taken as constants, the rate of change of temperature with time can be written as, dT -a (T - To) The above equation explains that, as the time increases, the difference in temperatures of the body and surroundings decreases and hence, the rate of fall of temperature also decreases. Page | 408 40 9 41 10 42 11 43 12 44 13 45 14 46 15 47 16 48 17 49 18 50 19 51 20 52 21 53 22 54 23 55 24 56 25 57 26 58 27 59 28 60 29 61 30 62 31 63 32 64 1. Plot the cooling curve, i.e, the graph of temperature (7) against time (), as shown in fig. 2 below: dT A .C/min B [2 time Figure 2: Graphical representation of results Figure 3: Treatment of results Page | 422. Choose five or six points on the cooling curve at regular intervals of say, T (50, 55, 60, 65, 70"C). Draw tangents to the cooling curve at these points and find the slope (gradients) dT / dt of the tangents. 3. Plot another graph of these slopes (gradients), dT / di, against temperature 7 at which these slopes are taken. Note that the graph of the slope d / diagainst 7 according to the equation above would be a straight line (fig.3). If this straight line is extended, it will cut the temperature axis at 7 because at this temperature d7 / di should be zero (no more change in temperature). Hence determine the room temperature (7, ) from the graph. 4. Using the two values of the room temperature obtained, calculate the % difference in T. 5. Water is heated to 80"C for 10 min. How much would be the temperature if k = 0.056 per min and the surrounding temperature is 25"C?Materials Required: Copper calorimeter Stirrer Wooden box with a clamp and stand . . . A wooden lid having a hole in the middle. Thermometer Stop watch Hot water of about 80 .C. . Simulator Procedure (as performed through the Online Labs): . Select Aluminium as the material of the calorimeter from the drop down list. . Select water as the liquid sample from the drop down list. . Select 75 C as the temperature of the preheated liquid using the slider. . Select 27 C as the room temperature using the slider. . Mass of the liquid is fixed as 250 g. . Click on the 'START' button on the timer to start/stop the experiment. . Record temperature after every 4 minutes, until time reaches 2HOURS:30MIN:00SEC . To redo the experiment, click on the 'Reset' button. PROCEDURE IS ALSO AVAILABLE ON VIDEO Constants Mass of liquid: 250 g Specific heat capacity of liquid: 4.19 KJ/KgK Specific heat capacity of calorimeter: 0.38 KJ/KgK Heat transfer co-efficient: 109 W/m.K Room temperature (7, ) = " C No. Time / Temperature No. Time / Temperature (min) T(C) (min) T(C) 1 33 2 34 3 35 4 36 5 37 6 38 39 Page | 41