When 1.60 multiplied by 105 J of heat enters a cherry pie initially at 20.0°C, its entropy increases by 470 J/K. What is its final temperature (C)?
I use delta S = entropy change = 1.6*10^5*[1/293 - 1/T2] = 470
Solve for T2, but it's not correct. Please someone help me to solve it. Thanks.
To solve for the final temperature (T2) of the cherry pie, we can use the formula:
ΔS = q / T
where:
- ΔS is the change in entropy (which is given as 470 J/K)
- q is the amount of heat transferred (which is given as 1.60 * 105 J)
- T is the temperature in Kelvin (initially 20°C + 273.15 to convert to Kelvin)
Plugging in the values, we have:
470 J/K = (1.60 * 105 J) / T2
To solve for T2, we rearrange the equation:
T2 = (1.60 * 105 J) / (470 J/K)
Simplifying, we have:
T2 ≈ 340.4 K
To convert this back to Celsius, subtract 273.15:
T2 ≈ 67.25°C
So, the final temperature of the cherry pie is approximately 67.25°C.