Heat: Solutions to Problems
-
a) 20C = (2 + 273) = 275 K
b) -132 0C = (132 + 273)= 141 K
-
a) 100 K = (100 - 273) 0C = -173 0C
b) 30 K = (30 - 273) = -2430C
c) 500 K = (500 - 273) = -243 0C
3. Given:
m = 0.700 kg
c = 920 J/KgK
t = 10 min = 600 s
T1 = 100 0C
T2 = 350C
m = 0.700 kg
c = 920 J/KgK
t = 10 min = 600 s
T1 = 100 0C
T2 = 350C
DT = T2 -T1 = (350C -100 0C)
= -65K
Find: the heat lost
Solution:
Q = m x c x DT
= (0.700 Kg)(920 J/KgK)(-65K)
= 41 860 J
= 42 KJ
4. Given:
V = 500 mL
m = 500 g = 0.500 kg
c = 4220 J/kgK
t = 4 min = 240 s
T2 = 100 0C
T1 = 200C
V = 500 mL
m = 500 g = 0.500 kg
c = 4220 J/kgK
t = 4 min = 240 s
T2 = 100 0C
T1 = 200C
DT = T2 -T1 = (100 0C-200C )
= 80K
Find:
-
the heat gained by the water
- the power of the kettle
Solution:
a) Q = m x c x DT
= (0.500 kg)(4200 J/kgK)(80K)
= 168 000 J
\ E = 168 000 J
b) P = E/t \ P = 168 000 J / 240 s
= 700 W
5. Given:
mcoffee = 200 g = 0.200 kg
mmilk = 15 g = 0.015 kg
clost = 4200 J/kgK *coffee
cgained = 3700 J/kgK *milk
T1 = 65 0C *coffee
T1 = 100C *milk
mcoffee = 200 g = 0.200 kg
mmilk = 15 g = 0.015 kg
clost = 4200 J/kgK *coffee
cgained = 3700 J/kgK *milk
T1 = 65 0C *coffee
T1 = 100C *milk
Find:
Final temperature of mixture (coffee + milk), let final temperature be T
Solution:
a) Qlost = Qgained
[m x c x DT]coffe = [m x c x DT]milk
(0.200 Kg)(4200 J/kgK)(65 0C-T)
= (0.015 kg)(3700 J/kgK)(T-10 0C)
\ 546 - 8.4T = 5.55T - 55
601 = 13.95T
Solve for T
\ T = 43.10C,
This is the final temperature