Question

A heater cool rated at 1000W is connected to a 110 V mains. How much time will take to melt 625 grams of ice at 0℃ (L= $80 {cal}/{gram}$).
A. 100 sec
B. 150 sec
C. 200 sec
D. 210 sec

Answer 1

Hint: To solve this problem, first find the heat produced by the heater in time t using the relationship between heat and power. Substitute the values in the formula. Then, find the amount of heat absorbed by ice to melt which shows the relationship between latent heat and heat absorbed. Then, substitute the given values. Equate both the equations and find the value of t. t is the amount of time taken to melt 625 grams of ice at 0℃.

Formula used:
$Q=Pt$
$Q=mL$

Complete answer:
Given: Power (P)= 1000 Watt
            Mass of ice= 625 grams= 0.625 kg
            Latent heat (L)= $80 {cal}/{gram}= 80 \times 4200= 336 \times {10}^{3} {J}/{kg}$
Let t be the time taken by ice to melt
Heat produced by heater in time t is given by,
$Q=Pt$
Substituting the value in above equation we get,
$Q=1000t$...(1)
Now, heat absorbed by ice to melt is given by,
$Q= mL$
Substituting the values in above equation we get,
$Q= 0.625 \times 336 \times {10}^{3}$
$\Rightarrow Q= 210 \times {10}^{3}$. ..(2)
Equating equation. (1) and (2) we get,
$1000t= 210 \times {10}^{3}$
$\therefore t= 210 s$
Hence, it will take 210 sec to melt 625 grams at 0℃.

So, the correct answer is “Option D”.

Note:
Before solving this question, students must take care of the units. They should convert the units either to CGS or MKS. 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.