I don't understand this question...
What is the change in entropy for a system that has 250 J of heat added to it at 25°C?
(Points : 1)
10 J/K
0.84 J/K
0.84 J/K
10 J/K
To determine the change in entropy for a system, we can use the formula:
ΔS = Q / T
where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 25°C + 273.15 = 298.15 K
Now, we can substitute the values into the formula:
ΔS = 250 J / 298.15 K
Calculating this, we get:
ΔS ≈ 0.837 J/K
Therefore, the change in entropy for the system is approximately 0.837 J/K.
Therefore, the correct answer is:
0.84 J/K
To calculate the change in entropy for a system, we can use the formula:
ΔS = Q / T
where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature of the system.
In this question, we are given that 250 J of heat is added to the system at a temperature of 25°C. To use the formula, we need to convert the temperature to Kelvin by adding 273.15 to it. So, 25°C + 273.15 = 298.15 K.
Now, we can substitute the values into the formula:
ΔS = 250 J / 298.15 K
Calculating this, we find:
ΔS ≈ 0.8378 J/K
Rounding to the nearest hundredth, the change in entropy is approximately 0.84 J/K.
Therefore, the correct answer is: 0.84 J/K.