Question

An office worker uses an immersion heater to warm 235 g of water in a light, covered, insulated cup from 20°C to 100°C in 5.00 minutes. The heater is a Nichrome resistance wire connected to a 120-V power supply. Assume that the wire is at 100°C throughout the 5.00-min time interval. (a) Calculate the average power required to warm the water to 100°C in 5.00 min. (The specific heat of water is 4186 J/kg · °C.)(Answer in units of Watts) (b) Calculate the required resistance in the heating element at 100°C.(Answer in ohms) (c) Calculate the resistance of the heating element at 20.0°C. (Answer in ohms) (d) Derive a relationship between the diameter of the wire, the resistivity at 20.0°C, ρ0, the resistance at 20.0°C, R0 , and the length L. (Do not substitute numerical values; use variables only.) (e) If L = 3.00 m, what is the diameter of the wire?(Answer in mm)

Answer 1

Part A.

Energy required to heat the water from 20 C to 100 C will be given by:

Q = m*C*dT

m = mass of water = 235 gm = 0.235 kg

C = Specific heat capacity of water = 4186 J/kg-C

dT = Change in temperature = 100 - 20 = 80 C

So,

Q = 0.235*4186*80 = 78696.8 J

Now Power provided by heater should be

Power = Energy required/time interval

time interval = 5.00 min = 300 sec

So,

Power = 78696.8 J/300 sec = 262.323 W = 262 W

Part B.

Now Since Power supply is done at 120 V, So Current through heater will be:

P = V*I

I = P/V = 262.323/120 = 2.186 Amp

Now From Ohm's law:

V = I*R

R = Resistance of wire = V/I

R = 120/2.186 = 54.895 ohm = 54.9 ohm

Part C.

Now Since wire is at 100 C during this time interval, So resistance of wire at 20 C will be:

R = R0*(1 + alpha*(Tf - Ti))

R0 = R/(1 + alpha*(Tf - Ti))

alpha = temperature coefficient of Nichrome = 0.4*10^-3

R0 = 54.895/(1 + 0.4*10^-3*(100 - 20)) = 53.193 Ohm

R0 = Temperature of nichrome at 20 C = 53.2 ohm

Part D.

Now we know that resistance of a wire is given by:

R0 = rho*L/A

rho = resistivity of nichrome, = 1.50*10^-6 Ohm-m

Suppose length of wire is L and diameter of wire is d, then

A = Cross-section Area of wire = pi*d^2/4

R0 = *L/(pi*d^2/4)

d = sqrt (4**L/(pi*R₀))

Part E.

Now when length of wire is L = 3.00 m, then

d = sqrt (4*1.50*10^-6*3.00/(pi*53.2))

d = 0.328*10^-3 m

d = 0.328 mm = diameter of wire

Let me know if you've any query.