Answer : The temperature of boiling water of part A (in Kelvin) = 373.15 K
The temperature of boiling water of part B (in Kelvin) = 372.15 K
The final temperature of cooling water of part A (in Kelvin) = 296.15 K
The final temperature of cooling water of part B (in Kelvin) = 280.15 K
Solution : Given,
The temperature of boiling water of part A = [tex]100^{0}C[/tex]
The temperature of boiling water of part B = [tex]99^{0}C[/tex]
The final temperature of cooling water of part A = [tex]23^{0}C[/tex]
The final temperature of cooling water of part B = [tex]7^{0}C[/tex]
Conversion of [tex]^{0}Celsius[/tex] into Kelvin.
[tex]T(K)=T(^{0}C)+273.15[/tex]
By using this conversion, we get
The temperature of boiling water of part A = [tex]100^{0}C[/tex] = 100 + 273.15 = 373.15 K
The temperature of boiling water of part B = [tex]99^{0}C[/tex] = 99 + 273.15 = 372.15 K
The final temperature of cooling water of part A = [tex]23^{0}C[/tex] = 23 + 273.15 = 296.15 K
The final temperature of cooling water of part B = [tex]7^{0}C[/tex] = 7 + 273.14 = 280.15 K
