Growth and decay (1 Viewer)

[atakach99]

An iceblock with temperature -14 degrees Celsius is left out in the sun. The air temperature is a constant 25 degrees Celsius and after 40 seconds the temperature of the iceblock has reached -5 degrees Celsius. Find

a) its temperature after 2 minutes

b) when the iceblock will start to melt (i.e when its temperature will reach 0 degrees Celsius.

plz help me solve this
thnks

 

[vds700]

An iceblock with temperature -14 degrees Celsius is left out in the sun. The air temperature is a constant 25 degrees Celsius and after 40 seconds the temperature of the iceblock has reached -5 degrees Celsius. Find

a) its temperature after 2 minutes

b) when the iceblock will start to melt (i.e when its temperature will reach 0 degrees Celsius.

plz help me solve this
thnks

Hint use Newtn's law of cooling:

T = 25 + Ae^Kt

Sub in given values to evaluate the constants, and u should be able to work the rest out.

 

[12o9]

Hint use Newtn's law of cooling:

T = 25 + Ae^Kt

Sub in given values to evaluate the constants, and u should be able to work the rest out.

are we required to remember that formula? beucase my school teacher hasnt even mentioned it ..

 

[lyounamu]

are we required to remember that formula? beucase my school teacher hasnt even mentioned it ..

No,you usually derive it yourself.

 

[vds700]

are we required to remember that formula? beucase my school teacher hasnt even mentioned it ..

As Namu said, tu need tro derive it. Its not hard. You start with dT/dt = k(T - P)

The, dt/dT = 1/k(T - P)
t = Int(1/k(T - P)dT
integrate and express T in terms of t and u shouild get T = P + Ae^kt

Often in exams, they say show that T = P + Ae^kt is a solution to dT/dt = k(T - P), and u just need to differentiate.