Question
Question: A metal ball cools from \({{64}^{\circ }}C\) to \({{50}^{\circ }}C\) in 10 minutes and to \({{42}^{\...
A metal ball cools from 64∘C to 50∘C in 10 minutes and to 42∘C in next 10minutes. The ratio of rates of fall of temperature during the two intervals is
A. 74
B. 47
C. 2
D. 2.5
Solution
Temperature is the degree of measuring the hotness or coldness of a body. When a body cools, its temperature reduces. Suppose the temperature reduces from T1 to T2 in time t, then the rate of cooling is equal to r=tT1−T2.
Complete step by step answer:
Let us first understand what is meant by the term temperature. Temperature is one of the fundamental physical quantities in nature. When we say that the temperature of a body is high, for example the temperature of the atmosphere is 40∘C, we mean that the body or the substance is hot. When we say that the temperature of a body is low, we mean that it is cold. We can consider ice as an example. Therefore, temperature is the degree of measuring the hotness or coldness of a body.
Hence, when a body cools, its temperature reduces. Suppose the temperature reduces from T1 to T2 in time t, then the rate of cooling is equal to r=tT1−T2.
It is given that the metal ball cools from 64∘C to 50∘C in 10 minutes and to 42∘C in next 10minutes.
This means that the first rate of cooling is r1=1064−50=1014=1.4∘C(min)−1 ….. (i)
And the second rate of cooling is r2=1050−42=108=0.8∘C(min)−1 …… (ii)
Now, divide (i) by (ii).
r2r1=0.81.4 ∴r2r1=47
This means that the ratio of rates of fall of temperature during the two intervals is 47.
Hence, the correct option is B.
Note: Students may misunderstand between heat and temperature. We already know that temperature tells us about the hotness and the coldness of a body. Heat is the energy exchanged between two bodies that are different temperatures.