In a 25°C room, hot coffee in a vacuum flask cools from 72°C to 48°C in eight hours. What will its temperature

Question

In a 25°C room, hot coffee in a vacuum flask cools from 72°C to 48°C in eight hours. What will its temperature be after another eight hours?

Answer 1

This is vacuum between 72°C and the room ambient 25°C, therefore the heat loss can never take place through conduction and convection processes, but the Del T the fall in temperature can surely take place upon radiatives losses - which do not require medium - as is absent.

Let the outer-wall temperature of flask (boundary before vacuum) having heat source of 72°C be at equilibrium with overall 72°C of the coffee. Also, let the outer-most-wall temperature of flask (boundary after vacuum) be at room temperature 25°C. Though, in real there will be intermittent temperature gradients as nothing equates as vacuum or perfect insulator.

Radiatives losses will take place between source at 72°C and 25°C (asymptotically).

Q-dot = sigma* A * [T^4 - To^4] where T is in deg K
or
m cp dT/dt = sigma* A * [T^4 - To^4]

m cp [T - T'] / dt = sigma* A * [T^4 - To^4]

dt = m cp [T - T1] / sigma* A * [T^4 - To^4]

Ist case
8 = m cp [345 - 321] / sigma* A * [(345)^4 - (298)^4] ----(1)

2nd case

8 = m cp [321 - T2] / sigma* A * [(321)^4 - (298)^4] ----(2)

dividing (1) by (2)

.......[345 - 321]...[(321)^4 - (298)^4]
1.=.----------------*-------------------------
.......[321 - T2].....[(345)^4 - (298)^4]

[321 - T2] = 24*[2731297265 / 6280800209]

[321 - T2] = 10.436

T2 = 321 - 10.436 = 310.56 deg K

T2 = 37.56 deg C after another 8 hours

Answer 2

The Ultimate Answer Guaranteed had the correct answer. Mathematically the temperature would drop to 24 after another 8 hrs but since it cannot be colder than the room temperature it will be 25°C.

Answer 3

Using Newton's Law of Cooling which is defined as:

dT/dt = -r(T - Te) where r is a constant>0 and Te is the environmental temperature (25 in this case).

dT/dt = -r(T-Te)
dT/(T-Te) = -rdt
int(dT/(T-Te) = -int(rdt)
ln|T-Te| = -rt+C
T-Te = C*e^(-rt)
Let T = T(0):
T(0) - Te = C*e^(-r0)
C = T(0) - Te

Therefore:
T = Te + (T(0) - Te)*e^(-rt)
From the problem:
T(0) = 72
T(8) = 48
Te = 25
48 = 25 + (72 - 25)*e^(-8r)
23 = 47*e^(-8r)
.489 ~= e^(-8r)
ln(.489) ~= -8r
-.715 ~= -8r
r ~= .0893

After a further eight hours:
T(0) = 48
Te = 25
t = 8

T ~= 25 + (48 - 25)*e^(-.0893*8)
T ~= 25 + 23*e^(-.7147)
T ~= 25 + 23*0.489
T ~= 25 +11.255
T ~= 36.255

Therefore, after a further eight hours the temperature will be approximately 36.26 degrees C.

Answer 4

Trick question:
25°C
because it cant be less than room temperature so once it hits 25 it stays the same temperature.

Answer 5

24 degrees C