A thermometer reading 75^{\circ}F is taken out where t

Joan Thompson

Joan Thompson

Answered question

2021-12-27

A thermometer reading 75F is taken out where the temperature is 20F. The reading is 30F 4 min later.
A. What is the value of k in four decimal places?
B. Find the thermometer reading 7 min after the thermometer was brought outside.
C. Find the time taken for the reading to drop from 75F to within a half degree of the air temperature.

Answer & Explanation

boronganfh

boronganfh

Beginner2021-12-28Added 33 answers

Step 1
Given that:
- Initial temperature =75F
- Surrounding temperature =20F
- The temperature is 30F after 4 minute
Step 2
According to the Newtons
ambarakaq8

ambarakaq8

Beginner2021-12-29Added 31 answers

Solution:
dTdt=k(TTs)
dTdt=k(T20)
dTT20=k dt
ln(T20)=kt+lnC
ln(T20)=lnekt+lnC
ln(T20)=lnCekt
T20=Cekt
T=20+Cekt
When t=0,T=75
75=20+Ce0
C=55
When t=4,T=30
30=20+55e4k
1055=e4k
ek=(211)14
Thus, T=20+55(211)t4
When T=20.5F
20.5=20+55(211)t4
0.555=(211)t4
ln0.555=ln(211)t4
ln(1110)=14tln(211)
t=4ln(1110)ln(211)
t=11.029 minutes
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

dTdt=k(TTs)dTdt=k(TTs)dTTTs=kdtdTTTs=kdtln(TTs)=kt+lncTTs=cektT=cekt+TsInitialy, t=0,and T=T0T0=ce0+TsT0Ts=cT(t)=Ts+(T0Ts)ektPlug T0=75,Ts=20 in aboveT(t)=20+(7520)ektT(t)=20+55ektNow plug t=4 and T=30 in above30=20+55e4ke4k=211k=ln(211)4k=0.4262T(t)=20+55e0.4262tPlug t=7 in aboveT(7)=20+55e0.4262(7)T(7)=22.7875F to 74.5FdTdt=k(TTs)dTdt=0.4262(T20)Plug dT=0.5 and T=75 in above0.5dt=0.4262(7520)dt=0.02133

user_27qwe

user_27qwe

Skilled2023-05-14Added 375 answers

Result:
t12.3413 minutes
Solution:
A. To solve for the value of k, we can use the formula for exponential decay:
T(t)=T0ekt,
where T(t) is the temperature at time t, T0 is the initial temperature, and k is the decay constant.
Given that the initial temperature T0 is 75 degrees F and the temperature at time t = 4 minutes is 30 degrees F, we have:
30=75e4k.
To solve for k, we can divide both sides by 75 and take the natural logarithm of both sides:
ln(3075)=4k.
Simplifying further:
ln(25)=4k.
Finally, we can solve for k by dividing both sides by -4:
k=ln(25)4.
Thus, the value of k, rounded to four decimal places, is approximately k0.2835.
B. To find the thermometer reading 7 minutes after it was brought outside, we can use the same formula as before:
T(t)=T0ekt.
Given that the initial temperature T0 is 75 degrees F and t = 7 minutes, we can substitute these values into the formula:
T(7)=75e0.2835·7.
Evaluating this expression, we find:
T(7)=75e1.9845.
Thus, the thermometer reading 7 minutes after it was brought outside is approximately T(7)27.2571 degrees F.
C. To find the time taken for the reading to drop from 75 degrees F to within a half degree of the air temperature, we need to determine the time it takes for the temperature to reach a value of 75 - 0.5 = 74.5 degrees F.
Using the formula T(t)=T0ekt and substituting T0=75 and T(t)=74.5, we have:
74.5=75ekt.
Dividing both sides by 75 and taking the natural logarithm of both sides, we get:
ln(74.575)=kt.
Simplifying further:
ln(298300)=kt.
Finally, we can solve for t by dividing both sides by -k:
t=ln(300298)k.
Using the previously calculated value of k, we can substitute it into the equation and evaluate t:
t=ln(300298)0.2835.
Thus, the time taken for the reading to drop from 75 degrees F to within a half degree of the air temperature is approximately t12.3413 minutes.
karton

karton

Expert2023-05-14Added 613 answers

A. To find the value of k, we can use the exponential decay formula:
T(t)=T0+(TaT0)·ekt,
where:
- T(t) is the temperature at time t,
- T0 is the initial temperature of the thermometer (75°F),
- Ta is the ambient temperature (20°F),
- k is the decay constant, and
- t is the time elapsed.
We are given that the temperature dropped to 30°F after 4 minutes. Plugging in the values:
30=75+(2075)·ek·4.
To solve for k, we can rearrange the equation as follows:
30752075=e4k.
Taking the natural logarithm of both sides, we get:
ln(30752075)=4k.
Now we can solve for k:
k=ln(30752075)4.
Calculating the value of k using a calculator or computer program, we find:
k0.0784 (rounded to four decimal places).
Therefore, the value of k is approximately 0.0784.
B. To find the thermometer reading 7 minutes after it was brought outside, we can use the same exponential decay formula:
T(t)=T0+(TaT0)·ekt.
Plugging in the values:
T(7)=75+(2075)·e0.0784·7.
Calculating this using a calculator or computer program, we find:
T(7)35.5073 (rounded to four decimal places).
Therefore, the thermometer reading 7 minutes after it was brought outside is approximately 35.5073°F.
C. To find the time taken for the reading to drop from 75°F to within a half degree of the air temperature, we need to solve the following equation:
T(t)=Ta+T0Ta2.
Plugging in the values:
T(t)=20+75202.
Simplifying the equation:
T(t)=47.5.
Now we can solve for t:
47.5=75+(2075)·e0.0784·t.
Rearranging the equation:
e0.0784·t=47.5752075.
Taking the natural logarithm of both sides:
0.0784·t=ln(47.5752075).
Solving for t:
t=ln(47.5752075)0.0784.
Calculating the value of t using a calculator or computer program, we find:
t8.5614 (rounded to four decimal places).
Therefore, the time taken for the reading to drop from 75°F to within a half degree of the air temperature is approximately 8

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