An experiment takes 10 minutes to raise the temperature of water in a container from 0°C to 100°C and another 55 minutes to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking specific heat of water to be $1{cal}/{g°C}$, the heat of vaporization according to this experiment will come out to be
A. 530 ${cal}/{g}$
B. 550 ${cal}/{g}$
C. 540 ${cal}/{g}$
D. 560 ${cal}/{g}$
Answer
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Hint: To solve this problem, first find the amount of heat supplied in 10 minutes. Then, find the amount of heat required to raise the temperature by 100°C. Equate both the equations and find the value of heat in terms of mass. Then, find the heat supplied in the next 55 minutes. Use the formula for latent heat and equate it with the equation for heat supplied in 55 minutes. Then, substitute the value of heat obtained above in terms of mass. Thus, calculate L which is the heat of vaporization.
Formula used: $Q= Pt$
${Q}_{1}= mL$
$Q= ms \Delta T$
Complete step by step answer:
Given: Specific heat of water (s)= $1{cal}/{g°C}$
${t}_{1}$= 10 mins= 600 secs
${t}_{2}$= 55 mins= 3300 secs
$\Delta T$= 100-0=$ 100°C$
Let the input power be P
The formula for heat supplied is given by,
$Q= Pt$ …(1}
Using equation. (1), heat supplied in 10 mins will be given by,
$Q= 600P$ …(2)
Heat required to raise the temperature by 100°C is given by,
$Q= ms \Delta T$
Substituting values in above equation we get,
$Q= m \times 100$
$\Rightarrow Q= 100m$ …(3)
Comparing equation. (2) and (3) we get,
$600P= 100m$
$\Rightarrow 6P=m$ …(4}
Now, heat supplied in the next 55 minutes is given by,
${Q}_{1}= 3300P$ …(5)
The formula for latent heat is given by,
${Q}_{1}= mL$ …(6)
Where, m is the mass
L is the specific latent heat
Comparing equation. (5) and (6) we get,
$3300P= mL$
Substituting equation. (4) in above equation we get,
$3300P= 6PL$
$\Rightarrow L= \dfrac {3300}{6}$
$\Rightarrow L= 550 {cal}/{g}$
Hence, according to this experiment, heat of vaporization will come out to be $550 {cal}/{g}$.
So, the correct answer is “Option B”.
Note: All phase changes release or absorb latent heat. There are three basic types of latent each associated with a different type of phase. When the phase changes from solid to liquid, the latent heat associated with it is called latent heat of fusion while the phase changes from liquid to gas, the latent heat associated with it is called latent heat of vaporization. The latent heat associated with the phase change from solid to gas is called latent heat of sublimation.
Formula used: $Q= Pt$
${Q}_{1}= mL$
$Q= ms \Delta T$
Complete step by step answer:
Given: Specific heat of water (s)= $1{cal}/{g°C}$
${t}_{1}$= 10 mins= 600 secs
${t}_{2}$= 55 mins= 3300 secs
$\Delta T$= 100-0=$ 100°C$
Let the input power be P
The formula for heat supplied is given by,
$Q= Pt$ …(1}
Using equation. (1), heat supplied in 10 mins will be given by,
$Q= 600P$ …(2)
Heat required to raise the temperature by 100°C is given by,
$Q= ms \Delta T$
Substituting values in above equation we get,
$Q= m \times 100$
$\Rightarrow Q= 100m$ …(3)
Comparing equation. (2) and (3) we get,
$600P= 100m$
$\Rightarrow 6P=m$ …(4}
Now, heat supplied in the next 55 minutes is given by,
${Q}_{1}= 3300P$ …(5)
The formula for latent heat is given by,
${Q}_{1}= mL$ …(6)
Where, m is the mass
L is the specific latent heat
Comparing equation. (5) and (6) we get,
$3300P= mL$
Substituting equation. (4) in above equation we get,
$3300P= 6PL$
$\Rightarrow L= \dfrac {3300}{6}$
$\Rightarrow L= 550 {cal}/{g}$
Hence, according to this experiment, heat of vaporization will come out to be $550 {cal}/{g}$.
So, the correct answer is “Option B”.
Note: All phase changes release or absorb latent heat. There are three basic types of latent each associated with a different type of phase. When the phase changes from solid to liquid, the latent heat associated with it is called latent heat of fusion while the phase changes from liquid to gas, the latent heat associated with it is called latent heat of vaporization. The latent heat associated with the phase change from solid to gas is called latent heat of sublimation.
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