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?
To calculate the change in entropy of the resistor, we need to know the change in temperature and the heat transfer involved. The change in temperature can be found using the specific heat capacity formula:
Q = mcΔT
Where Q is the heat transfer, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
The heat transfer, Q, can be calculated using the formula:
Q = I^2Rt
Where I is the current, R is the resistance, and t is the time.
Given that the current is 1A, the resistance is 50Ω, and the time is 1s, we can calculate Q as follows:
Q = (1A)^2 * 50Ω * 1s = 50 Joules
Since we know the mass is 5g, the specific heat capacity is 850 J/kg/K, and the initial temperature is 10 degrees, we can calculate the change in temperature (ΔT) using the specific heat capacity formula:
50J = (5g) * (850 J/kg/K) * ΔT
ΔT = (50J) / (5g * 850 J/kg/K) ≈ 0.0118 K
The change in entropy (ΔS) of the resistor can be calculated using the formula:
ΔS = Q / T
Where T is the temperature in Kelvin. Since the change in temperature (ΔT) is in Kelvin, we can use the initial temperature to calculate T:
T = 10°C + ΔT = 10K + 0.0118K ≈ 10.0118 K
Now we can calculate the change in entropy of the resistor:
ΔS = 50J / 10.0118 K ≈ 4.996 J/K
To calculate the change in entropy of the universe, we need to consider both the change in entropy of the resistor and the surroundings (which includes the rest of the universe). In this case, the surroundings can be considered a thermal reservoir due to it being well-insulated.
The change in entropy of the universe (ΔS_universe) can be calculated using the equation:
ΔS_universe = ΔS_resistor + ΔS_surroundings
Since the surroundings are a thermal reservoir, the change in entropy of the surroundings can be calculated using the formula:
ΔS_surroundings = -Q_surroundings / T_surroundings
Since the resistor is thermally insulated, there is no heat transfer between the resistor and the surroundings, so Q_surroundings is equal to zero. This means that the change in entropy of the surroundings (ΔS_surroundings) is also zero. Therefore, the change in entropy of the universe (ΔS_universe) is simply equal to the change in entropy of the resistor:
ΔS_universe = ΔS_resistor ≈ 4.996 J/K
Thus, the change in entropy of the resistor is approximately 4.996 J/K, and the change in entropy of the universe is also approximately 4.996 J/K.