A thermally insulated 50 ohm resistor carries a current of 1 A for 1 s. The initial temperature of the resistor is 10 degree Celsius, its mass is 5 g, its specific heat capacity is 850. What is the change in entropy of the resistor?

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

ΔS = m * C * ln(Tf/Ti)

Where:
ΔS = Change in entropy
m = Mass of the resistor
C = Specific heat capacity of the resistor
Tf = Final temperature of the resistor
Ti = Initial temperature of the resistor

Given:
m = 5 g
C = 850
Ti = 10 °C

To find Tf, we can use the formula:

Q = m * C * Δθ

Where:
Q = Heat absorbed by the resistor
m = Mass of the resistor
C = Specific heat capacity of the resistor
Δθ = Change in temperature

Rearranging the equation, we get:

Δθ = Q / (m * C)

We know that:
Q = I^2 * R * t

Where:
Q = Heat absorbed by the resistor
I = Current flowing through the resistor
R = Resistance of the resistor
t = Time

Given:
I = 1 A
R = 50 Ω
t = 1 s

Plugging in the values, we get:

Q = (1 A)^2 * (50 Ω) * (1 s)
Q = 50 J

Now, we can calculate Δθ:

Δθ = (50 J) / (5 g * 850)
Δθ ≈ 0.0118 °C

Finally, we can calculate the final temperature Tf:

Tf = Ti + Δθ
Tf = 10 °C + 0.0118 °C
Tf ≈ 10.0118 °C

Now we can calculate the change in entropy ΔS:

ΔS = (5 g) * (850) * ln(10.0118 °C / 10 °C)
ΔS ≈ 5 g * 850 * ln(1.00118)
ΔS ≈ 5 g * 850 * 0.00118
ΔS ≈ 5.043 J/K

Therefore, the change in entropy of the resistor is approximately 5.043 J/K.

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

ΔS = m * c * ln(Tf / Ti)

Where:
ΔS is the change in entropy,
m is the mass of the resistor,
c is the specific heat capacity of the resistor,
Tf is the final temperature of the resistor, and
Ti is the initial temperature of the resistor.

Given:
m = 5 g (convert to kg: 5 g = 0.005 kg),
c = 850 J/kg°C,
Ti = 10°C.

First, we need to calculate the final temperature Tf. To do this, we can use the formula:

Q = I^2 * R * t

Where:
Q is the heat generated by the resistor,
I is the current,
R is the resistance, and
t is the time.

Given:
I = 1 A,
R = 50 ohms,
t = 1 s.

Substituting the values, we get:

Q = (1 A)^2 * 50 ohms * 1 s
= 50 J

Since the resistor is thermally insulated, all the heat generated is retained within the resistor.

Next, let's calculate the change in temperature (ΔT) using the formula:

ΔT = Q / (m * c)

Substituting the values, we get:

ΔT = 50 J / (0.005 kg * 850 J/kg°C)
≈ 11.76 °C

Now, let's find the final temperature Tf:

Tf = Ti + ΔT
= 10°C + 11.76°C
≈ 21.76°C

Finally, we can substitute all the values into the entropy formula to find the change in entropy (ΔS):

ΔS = m * c * ln(Tf / Ti)
= 0.005 kg * 850 J/kg°C * ln(21.76°C / 10°C)
≈ 0.246 J/°C

Therefore, the change in entropy of the resistor is approximately 0.246 J/°C.

First calculate how much the temperature rises. Neglect heat loss during the brief interval of resistive heating. The deposited heat energy is

Q = I^2*R*t = 50 Joules.

You have not said what your units of heat capacity (C) are, so I cannot tell you what the temperature rise is. The formula to use is:
(T2 - T1) = Q/(M*C)

The entropy change is the integral of dQ/T, which is

M*C*ln(T2/T1)

where T1 and T2 are initial and final absolute temperatures.

Thank you- this is very helpful

Well, talking about entropy, it's a bit like trying to explain the concept of humor to a rock. But hey, I'll give it a shot.

Entropy is a measure of the disorder or randomness of a system. In this case, since we're dealing with a thermally insulated resistor, we can assume that it's an isolated system.

Now, the change in entropy (ΔS) can be calculated using the formula ΔS = Q/T, where Q is the heat gained or lost by the system, and T is the temperature.

Since the resistor is thermally insulated, there is no heat transfer in or out of the system, so Q is zero. Therefore, we can confidently say that the change in entropy of the resistor is... zero!

I know, I know, it's not the most exciting answer. But hey, at least we didn't go off on a tangent about why resistors don't go to parties, right?