A kettle full of boiling water was brought to a boil in a room with temperature

Question

A kettle full of boiling water was brought to a boil in a room with temperature 20 degree C, and then all heating of the water ceased. After exactly 7 minutes and 30 seconds the temperature of the water had decreased from 100 degree C to 75 degree C. (a) Construct a model of this system using Newton's Law of Cooling in the form T(t) = T_s + (T_0 - T_s) e^-kt where t is time measured in minutes, T_0 is the initial temperature of the water in the kettle (in degrees Celsius), T_s is the ambient room temperature (in degrees Celsius), T(t) is the water temperature (in degrees Celsius) at time t and k is a rate constant. Here constructing the model involves substituting the initial water temperature and the ambient room temperature into the given equation for Newton's Law of Cooling, and then simplifying the equation where possible. (b) Hence determine the rate constant, k. (c) Determine the temperature of the water another 10 minutes after the temperature reading of 75 degree C. A hair sample from a Coptic mummy was found to contain 82% of the relative proportion of carbon-14 found in living matter and the atmosphere. Approximately how long before the carbon-14 measurements were made had the mummy been buried? (The half-life of carbon-14 is 5730 years.) The most desctructive earthquake in contemporary Australian history occurred in Newcastle, NSW, in 1989. The earthquake had a magnitude of 5.6. (a) Assuming seismograph readings were taken 100km from the epicentre of the earthquake, what was the corresponding intensity of the earthquake (measured by the amplitude of a seismograph reading in centimetres)? (b) The 1960 Valdivia earthquake was the most powerful earthquake ever recorded. Its approximate intensity was 3.1623 times 10^5 cm. What was the magnitude of this earthquake on the Richter scale? (c) How many times more intense than the 1989 Newcastle earthquake was the Valdivia earthquake?

Explanation / Answer

Question-1:

Given

Surrounding Temperature, Ts = 20o C

At time t=0 sec, Temperature of water,T0 = 100o C

At time t= 450 sec (7 min 30 sec) ,Temperature of water,T = 75o C

a.

Newton's Law of Cooling Equation, T(t) = Ts + (T0-Ts)e-kt

Substituting above data in above equation we get,

T(t) = 20 + (100-20)e-kt

T(t) = 20 + 80e-kt , t is in sec

b.

Now from given data, At t= 450 sec, T(450) = 75oC

Substituting above equation, we get,

T(450) = 20 + 80e-450k

75= 20 + 80e-450k

Hence, e-450k = 55/80

Taking log on both sides, we get, -450k=ln(55/80)

-450k=ln(0.6875)

-450k=-0.3747

Hence, k = 0.000833 sec-1

c.

Here, For 450 sec, the temperature reading is 75oC, Now we need to calculate tempearature after 10 min of reaching 75oC. This means the time from intial temperature would be, t = 450 + 10*60 = 1050 sec.

From Newton law of cooling equation

T(t) = 20 + 80e-0.000833t

T(1050) = 20 + 80e-0.000833*1050

T(1050) = 20 + 80e-0.87465

T(1050) = 20 + 80*0.41707

T(1050) = 20 + 33.3656

Hence, T(1050) = 53.3656oC

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