Question

Solve the following examples.

a.What must be the temperature in Fahrenheit so that it will be twice its value in Celsius?

b. A bridge is made from 20 m long iron rods. At temperature 18^∘C, the distance between two rods in 0.4 cm. Up to what temperature will the bridge be in good shape?

c. At 15^∘C the height of Eifel tower is 324 m. If it is made of iron, what will be the increase in length in cm, at 30^∘C?

d. Two substances A and B have specific heats c and 2 c respectively. If A and B are given Q and 4Q amounts of heat respectively, the change in their temperatures is the same. If the mass of A is m, what is the mass of B?

e. When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by 10^∘C. What is the specific heat of the substance?

Answer 1

a. Let the temperature in Celsius be T.
So, the temperature in Fahrenheit = 2T
Now,
$$\begin{gathered} F=32+\frac{9}{5}\mathrm{C} \\ \Rightarrow 2T=32+\frac{9}{5}T \\ T={160}^{\mathrm{o}}\mathrm{C} \\ \Rightarrow \mathrm{Temperature}\mathrm{in}\mathrm{Fahrenheit}=2T=320{}^{\mathrm{o}}\mathrm{F} \end{gathered}$$

b. Length of the iron rod = 20 m = 2000 cm at 18^∘C
Distance between the length of two rods, $=0.4\mathrm{cm}$
Temperature coefficient of linear expansion of iron = $11.5\times {10}^{-6}{}^{\mathrm{o}}{\mathrm{C}}^{-1}$
The bridge will be in good shape till both the rods expand by 0.2 cm as the temperature is increased. Let at temperature T, both the rods expand by 0.2 cm i.e. the total expansion is 0.4 cm.
Using formula for linear expansion of solids, we have

$$\begin{gathered} \Rightarrow \frac{0.4}{2000}=11.5\times {10}^{-6}\times (T-16) \\ T=18+\frac{0.4}{2000\times 11.5\times {10}^{-6}}=35.4{}^{\mathrm{o}}\mathrm{C} \end{gathered}$$

c. Height of Eifel tower = 324 m = 32400 cm at 15^∘C
Temperature coefficient of linear expansion of iron = $11.5\times {10}^{-6}{}^{\mathrm{o}}{\mathrm{C}}^{-1}$
Change in temperature = 30^∘C $-$ 15 ^∘C = 15^∘C
Change in length = $∆l$
Using formula for linear expansion of solids, we have

$$\begin{gathered} \Rightarrow \frac{∆l}{32400}=11.5\times {10}^{-6}\times 15 \\ ∆l=32400\times 11.5\times {10}^{-6}\times 15=5.6\mathrm{cm} \end{gathered}$$

d. Let the mass of B M.
Let the change in temperature be T for both the bodies, A and B.
The amount of heat in a body is given as
$Q=m\times c\times ∆T$
For body A,
$$\begin{gathered} Q=m\times c\times T \\ \Rightarrow T=\frac{Q}{mc}.....\left(\mathrm{i}\right) \end{gathered}$$
For body B,
$$\begin{gathered} 4Q=M\times 2c\times T \\ \Rightarrow M=\frac{4Q}{2c\times T} \\ \mathrm{From}\left(\mathrm{i}\right),T=\frac{Q}{mc} \\ \Rightarrow M=\frac{4Q}{2c\times \frac{Q}{mc}}=2m \end{gathered}$$

e. Let the specific heat capacity of the substance be c.
Given: Mass of the substance, m = 3 kg = 3000 g
Heat given to the substance, Q = 600 cal
Increase in temperature of the substance = 10^∘C
Now, the amount of heat in a body is given as
$$\begin{gathered} Q=m\times c\times ∆T \\ \Rightarrow c=\frac{Q}{m\times ∆T}=\frac{600}{3000\times 10}=0.02\mathrm{cal}{\mathrm{g}}^{-1}{}^{\mathrm{o}}{\mathrm{C}}^{-1} \end{gathered}$$