A 10- Ω coil takes 21 s to melt 10g of ice at 0oC. Assuming no heat losses, determine the current in the coil. (Specific latent heat of fusion of ice = 336 ...
A 10- Ω coil takes 21 s to melt 10g of ice at 0oC. Assuming no heat losses, determine the current in the coil. (Specific latent heat of fusion of ice = 336 Jg-1)
Answer Details
The amount of heat required to melt 10g of ice is given by:
Q = mL
where Q is the heat required, m is the mass of ice and L is the specific latent heat of fusion of ice.
Therefore, Q = (10 g) x (336 Jg^-1) = 3360 J.
The time taken to melt the ice is 21 s.
The power output of the coil is given by:
P = Q/t
where P is the power output, Q is the heat required and t is the time taken.
Therefore, P = (3360 J)/(21 s) = 160 W.
Using Ohm's law, we can find the current in the coil:
V = IR
where V is the voltage, I is the current and R is the resistance.
The resistance of the coil is given as 10 Ω, and the voltage is not given directly in the question. However, we can use the power output to find the voltage:
P = IV
Therefore, V = P/I = (160 W)/(I).
Substituting this value of V in the equation V = IR, we get:
(I)R = (160 W)/(I)
Solving for I, we get:
I^2 = (160 W)/(10 Ω) = 16 A
Taking the square root of both sides, we get:
I = 4 A.
Therefore, the current in the coil is 4A.
Answer: 4A.