A portion of an electric circuit connected to a 40-ohm resistor is embedded in 0

Question

A portion of an electric circuit connected to a 40-ohm resistor is embedded in 0.20 kilogram of a solid substance in a calorimeter.The external portion of the circuit is connected to a 60-volt power supply, as shown above.
a) Calculate the current in the resistor.
b) Calculate the rate at which heat is generated in the resistor.
c) Assuming that all of the heat generated by the resistor is absorbed by the solid substance, and that it takes 4 minutes to raise the temperature of the substance from 20degrees c to 80degrees c, calculate the specific heat of the substance.
d) At 80 degrees c the substance begins to melt. The heat of fusion of the substance is 1.35 x 10^5 j/kg. HOw long after the temperature reaches 80 will it take to melt all of the substance?

Explanation / Answer

(a)

By V = i x R

i = V/R = 60/40 = 1.5 amp

(b)

By P = i^2 x R

P = (1.5)^2 x 40 = 90 J/sec

(c)

The energy given in 4 minute will be
H = i^2 x R x t
H = 90 x 4 x 60 = 21600 J
By Q = m x c x delta T
21600 = 0.2 x c x (80-20)
c = 1800 J/kg-*C

(d)

Energy required to melt the substance,
By Q = m x L
Q = 0.20 x 1.35 x 10^5 = 27000 J
Let the heat energy is produced in t sec
By H = i^2 x R x t
27000 = 90 x t
t = 300 sec = 5 minutes

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