
A metal ball cools from ${{64}^{\circ }}C$ to ${{50}^{\circ }}C$ in 10 minutes and to ${{42}^{\circ }}C$ in next 10minutes. The ratio of rates of fall of temperature during the two intervals is
A. $\dfrac{4}{7}$
B. $\dfrac{7}{4}$
C. $2$
D. $2.5$
Answer
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Hint: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 ${{T}_{1}}$ to ${{T}_{2}}$ in time t, then the rate of cooling is equal to $r=\dfrac{{{T}_{1}}-{{T}_{2}}}{t}$.
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 ${{40}^{\circ }}C$, 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 ${{T}_{1}}$ to ${{T}_{2}}$ in time t, then the rate of cooling is equal to $r=\dfrac{{{T}_{1}}-{{T}_{2}}}{t}$.
It is given that the metal ball cools from ${{64}^{\circ }}C$ to ${{50}^{\circ }}C$ in 10 minutes and to ${{42}^{\circ }}C$ in next 10minutes.
This means that the first rate of cooling is ${{r}_{1}}=\dfrac{64-50}{10}=\dfrac{14}{10}={{1.4}^{\circ }}C{{\left( \min \right)}^{-1}}$ ….. (i)
And the second rate of cooling is ${{r}_{2}}=\dfrac{50-42}{10}=\dfrac{8}{10}={{0.8}^{\circ }}C{{\left( \min \right)}^{-1}}$ …… (ii)
Now, divide (i) by (ii).
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1.4}{0.8}\\
\therefore\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{7}{4}$
This means that the ratio of rates of fall of temperature during the two intervals is $\dfrac{7}{4}$.
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.
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 ${{40}^{\circ }}C$, 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 ${{T}_{1}}$ to ${{T}_{2}}$ in time t, then the rate of cooling is equal to $r=\dfrac{{{T}_{1}}-{{T}_{2}}}{t}$.
It is given that the metal ball cools from ${{64}^{\circ }}C$ to ${{50}^{\circ }}C$ in 10 minutes and to ${{42}^{\circ }}C$ in next 10minutes.
This means that the first rate of cooling is ${{r}_{1}}=\dfrac{64-50}{10}=\dfrac{14}{10}={{1.4}^{\circ }}C{{\left( \min \right)}^{-1}}$ ….. (i)
And the second rate of cooling is ${{r}_{2}}=\dfrac{50-42}{10}=\dfrac{8}{10}={{0.8}^{\circ }}C{{\left( \min \right)}^{-1}}$ …… (ii)
Now, divide (i) by (ii).
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1.4}{0.8}\\
\therefore\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{7}{4}$
This means that the ratio of rates of fall of temperature during the two intervals is $\dfrac{7}{4}$.
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.
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