What amount of current would pass through a 10 Ω coil if it takes 21s for the coil to just melt a lump of ice of mass 10g at 0oC if there are no heat losses...
What amount of current would pass through a 10 Ω coil if it takes 21s for the coil to just melt a lump of ice of mass 10g at 0oC if there are no heat losses? (Latent heat of fusion of ice = 336 Jg-1).
Answer Details
We can solve this problem by first finding the amount of heat energy required to melt the ice and then using that to calculate the current passing through the coil. The amount of heat energy required to melt the ice is given by: Q = mL where Q is the amount of heat energy, m is the mass of the ice, and L is the latent heat of fusion of ice. Substituting the given values, we get: Q = 10g x 336 Jg-1 = 3360 J Since there are no heat losses, all the electrical energy supplied to the coil is converted into heat energy to melt the ice. The electrical energy supplied is given by: E = VIt where E is the electrical energy, V is the voltage, I is the current, and t is the time taken. Since the voltage is not given, we cannot directly solve for the current. However, we can use the fact that the coil melts the ice in 21 seconds to find the current. Since all the electrical energy supplied is used to melt the ice, we can equate E and Q: VIt = 3360 J Solving for I, we get: I = 3360 J / (Vt) Substituting the given values, we get: I = 3360 J / (10 Ω x 21 s) = 16.00 A Therefore, the amount of current passing through the coil is 16.00 A. Answer: 16.00A