Advertisements
Advertisements
Question
Solve the following problem.
A metal sphere cools at the rate of 0.05 ºC/s when its temperature is 70 ºC and at the rate of 0.025 ºC/s when its temperature is 50 ºC. Determine the temperature of the surroundings and find the rate of cooling when the temperature of the metal sphere is 40 ºC.
Advertisements
Solution
Given: T1 = 70 ºC, `("dT"/"dt")_1,` 0.05 ºC/s
T2 = 50 ºC, `("dT"/"dt")_2,` 0.025 ºC/s, T3 = 40 ºC.
To find: i. Temperature of surrounding (T0)
ii. Rate of cooling `("dT"/"dt")_3`
Formula: `"dT"/"dt" = "C"("T" - "T"_0)`
Calculation: From formula,
`("dT"/"dt")_1 = "C"("T"_1 - "T"_0) and ("dT"/"dt")_2 = "C"("T"_2 - "T"_0)`
∴ `0.05/0.025 = ("C"(70 - "T"_0))/("C"(50 - "T"_0))`
∴ 2 (50 - T0) = 70 - T0
∴ T0 = 30 πC
Substituting value of To,
0.05 = C(70 - 30)
∴ C = `0.05/40 = 0.00125`/s.
For T3 = 40 ºC
`("dT"/"dt")_3 = "C"("T"_3 - "T"_0) = 0.00125 (40 - 30)`
= 0.00125 × 10
= 0.0125 ºC/s.
- Temperature of surroundings is 30 ºC.
- Rate of cooling at 40 ºC is 0.0125 ºC/s.
APPEARS IN
RELATED QUESTIONS
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
On a hot summer day we want to cool our room by opening the refrigerator door and closing all the windows and doors. Will the process work?
On a cold winter night you are asked to sit on a chair. Would you like to choose a metal chair or a wooden chair? Both are kept in the same lawn and are at the same temperature.
The temperature of the atmosphere at a high altitude is around 500°C. Yet an animal there would freeze to death and not boil. Explain.
Newton's law of cooling is a special case of
A body cools down from 65°C to 60°C in minutes. It will cool down from 60°C to 55°C in
Answer the following question.
State Newton’s law of cooling and explain how it can be experimentally verified.
A metal sphere cools from 80 °C to 60 °C in 6 min. How much time with it take to cool from 60 °C to 40 °C if the room temperature is 30 °C?
Rate of cooling of a body is 0.4 °C/min when excess temperature is 20 °C. The proportionality constant is ______.
A metal sphere cools from 66° C to 57° C in 10 minutes and to 44° C in the next 10 minutes. The ratio of fall of temperature of first 10 minutes to next ten minutes is ____________.
A liquid with a certain surface area takes 10 minutes to cool from 80° C to 70° C. The time taken by it to cool from 80° C to 60° C is [The surrounding temperature being 40° C] ____________.
Newton's law of cooling leads to the expression:
A tub of hot water cools from 80°C to 75°C in time t1 from 75°C to 70°C in time t2, and from 70°C to 65°C in time t3 then:
A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken for a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is ______
A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is ______.
Is the bulb of a thermometer made of diathermic or adiabatic wall?
In 5 minutes, a body cools from 75°C to 65°C at a room temperature of 25°C. The temperature of the body at the end of the next 5 minutes is ______°C.
In \[\frac{dT}{dt}\] = C(T−T₀), what does the constant C represent?
A body cools from 60°C to 40°C in 6 minutes. After next 6 minutes its temperature will be (Temperature of the surroundings is 10°C) ______.
Why is the calorimeter kept in open air or a constant-temperature enclosure during Newton’s law of cooling experiment?
A body is cooling from 80°C to 60°C in a room at 30°C. Compared to the time taken to cool from 60°C to 40°C, the time taken to cool from 80°C to 60°C will be ______.
The cooling curve of a hot body is a graph of temperature T versus time t. What is the correct description of this curve?
