Help please due in 10 min
1 answer:
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Answer:
- 28.3 minutes
- 53.6°C
Step-by-step explanation:
The temperature of the soup is described by Newton's Law of Cooling, which says the rate of change of temperature is proportional to the difference between the temperatures of the soup and the room. The solution to this differential equation is the exponential function ...
f(t) = a +b·c^(t/τ)
where a is the room temperature, b is the initial difference in temperature, and c is the fractional change in the difference in temperature over time period τ.
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<h3>application</h3>For the given scenario, we find ...
a = room temperature = 20 . . . . degrees C
b = (180 -20) = 160 . . . . degrees C
c = (100 -20)/(180 -20) = 80/160 = 1/2 . . . . in τ = 20 minutes
So, the formula for the temperature of the soup is ...
f(t) = 20 +160(1/2)^(t/20) . . . . . . . degrees C after t minutes
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<h3>time to 80°C</h3>Solving for t when f(t) = 80, we find ...
80 = 20 +160(1/2)^(t/20)
3/8 = (1/2)^(t/20) . . . . . subtract 20, divide by 160
20×log(3/8)/log(1/2) = t ≈ 28.3 . . . take logarithms, divide by coefficient of t
It will take about 28.3 minutes to cool to 80°C.
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<h3>temp at 45 minutes</h3>The temperature after 45 minutes is ...
f(45) = 20 +160(1/2)^(45/20)
f(45) ≈ 53.6 . . . . degrees C
After 45 minutes the temperature of the soup will be about 53.6°C.
