A thermally insulated 50ohm resistor carries a current of 1A for 1s. The initial temperature of the resistor is 10degree, its mass is 5g, and its specific heat capacity is 850j/kg/k. Calculate the change in entropy of the resistor and what is the change in entropy of the universe?

Question

A thermally insulated 50ohm resistor carries a current of 1A for 1s. The initial temperature of the resistor is 10degree, its mass is 5g, and its specific heat capacity is 850j/kg/k. Calculate the change in entropy of the resistor and what is the change in entropy of the universe?

Answer 1

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Answer 2

To calculate the change in entropy of the resistor, we need to use the formula:

ΔS = mcΔT / T

Where:
ΔS = Change in entropy
m = Mass of the resistor
c = Specific heat capacity of the resistor
ΔT = Change in temperature
T = Initial temperature of the resistor

Given data:
m = 5g = 0.005kg (converting grams to kg)
c = 850 J/kg/K
ΔT = Final temperature - Initial temperature = T_f - T_i

Initial temperature, T_i = 10°C = 10 + 273.15 K (converting to Kelvin)
Let's assume that the final temperature, T_f, is obtained by passing the current through the resistor for 1 second.

Now, we need to calculate the final temperature, T_f.

To do this, we need to know the power dissipated by the resistor, which can be calculated using the formula:

P = IV

Where:
P = Power
I = Current
V = Voltage

Given data:
I = 1A
R = 50 Ω (resistance)

Using Ohm's Law: V = IR

V = 1A × 50 Ω = 50V

Therefore, the power dissipated by the resistor is 50W.

Now, we can calculate the energy dissipated by the resistor using the formula:

E = Pt

Where:
E = Energy
P = Power
t = Time

Given data:
t = 1s

E = 50W × 1s = 50J

Now, we can calculate the final temperature, T_f, using the energy dissipated, E:

E = mcΔT

ΔT = E / mc

ΔT = 50J / (0.005kg * 850 J/kg/K)

ΔT = 11.76 K

Now we have the change in temperature, ΔT = T_f - T_i.

ΔT = T_f - T_i
11.76 K = T_f - (10 + 273.15) K

T_f = 11.76 K + (10 + 273.15) K
T_f = 295.91 K

Now, we can calculate the change in entropy, ΔS:

ΔS = mcΔT / T
ΔS = (0.005kg)(850 J/kg/K)(11.76 K) / (10 + 273.15) K

ΔS = 0.0625 J/K

Now, to calculate the change in entropy of the universe, we need to consider both the change in entropy of the resistor and the change in entropy of the surroundings.

Since the resistor is thermally insulated, there is no heat transfer to the surroundings. Therefore, the change in entropy of the surroundings is zero.

Hence, the change in entropy of the universe is equal to the change in entropy of the resistor, which is ΔS = 0.0625 J/K.