Question

A coil of wire of resistance 50Ω is embedded in an exceedingly block of ice. If a possible distinction of 210 V is applied across the coil, then the quantity of ice fusible per second is
A. 2.625 g
B. 3.68 g
C. 4.12 g
D. 4.12 kg

Answer 1

Hint: First of all, determine the heat in the block of ice due to the coil of wire which has resistance \[50\Omega \]. The heat that we will get will be in joules. So, change it into calories. Because 80 calorie heat is required to convert 1 g of ice at \[{0^ \circ }C\] to 1 g of water at \[{0^ \circ }C\].

Formula used:
The expression of heat generated in the wire is,
\[H = {\dfrac{{{V^2}}}{R}} \]
Where, $V$ is the potential difference and $R$ is the resistance.

Complete step by step solution:
Let us assume that the heat generated in a coil of wire is H. Now we have a coil of wire which has a resistance of 50Ω and is embedded in a block of ice.
And, the potential difference(V) = 220V, and the resistance(R) = 50Ω.
And we have to determine the amount of ice melted per second.

Therefore, we know that when the voltage is applied to a conducting wire having the resistance (R), then the heat generated in a coil of wire is given as,
\[H = {\dfrac{{{V^2}}}{R}} \]
In this problem, we have value. Therefore, we will get
\[H = {\dfrac{{210 \times 210}}{{50}}} \]
\[ \Rightarrow H = {882\,joule} \]

Now, we know that 80 calorie heat is required to convert 1 g of ice at \[{0^ \circ }C\] to 1 g of water at \[{0^ \circ }C\]. Therefore, we will convert the heat into calories. We have,
\[{1joule} = {\dfrac{1}{{4.2}}calorie} \]
Therefore, the generated heat will be,
\[H = {\dfrac{{882}}{{4.2}}calorie} \]
\[ \Rightarrow H = {210\,calorie} \]

Now, we know the heat is directly proportional to the mass of the substance and the change in the temperature. Therefore, we will write it as,
\[ \Rightarrow {210} = {m \times 80} \]
\[ \Rightarrow m = {2.625\,g} \]
Now the final answer is 2.625 g.

Hence, the correct answer is option A.

Note: It is important to note that 80 calorie heat is required to convert 1 g of ice at \[{0^ \circ }C\] to 1 g of water at \[{0^ \circ }C\]. And units must be the same to evaluate the problem.