Respuesta :
How much heat is contained in 100 kg of water at 60.0 °C?
answer= 2.77 x 10^4 kcal
Calculate the energy needed to vaporize 10.0 kg of iron (C = 0.110) initially at 2,640°C.
Answer= 16,300
I hope this helps :)
answer= 2.77 x 10^4 kcal
Calculate the energy needed to vaporize 10.0 kg of iron (C = 0.110) initially at 2,640°C.
Answer= 16,300
I hope this helps :)
1.
Answer:
1300 k Cal
Explanation:
Total heat contained by ice at -23 degree C is given by
[tex]Q = ms\Delta T[/tex]
here we know that
[tex]m = 10 kg[/tex]
s = 500 Cal/kg K
[tex]\Delta T = (-23) - (-273)[/tex]
[tex]\Delta T = 250 [/tex]
now by above formula we have
[tex]Q = 10(500)(250)[/tex]
[tex]Q = 1300 kCal[/tex]
2
Answer:
[tex]2.77 \times 10^4 k Cal[/tex]
Explanation:
Total heat contained by water at 60 degree C is given by
[tex]Q = ms_i\Delta T + mL + ms_w\Delta T'[/tex]
here we know that
[tex]m = 100 kg[/tex]
[tex]s_w = 1000 Cal/kg K[/tex]
[tex]s_i = 500 cal/kg K[/tex]
[tex]\Delta T = 0 - (-273) = 273[/tex]
[tex]\Delta T' = (60) - (0)[/tex]
[tex]\Delta T' = 60 [/tex]
now by above formula we have
[tex]Q = 100(500)(273) + 100(80000) + 100(1000)(60)[/tex]
[tex]Q = 2.77\times 10^4 kCal[/tex]
3
Answer:
[tex]16,300 k Cal[/tex]
Explanation:
Energy required to vaporize the iron is given as
[tex]Q = ms\Delta T + mL[/tex]
[tex]m = 10 kg[/tex]
[tex]C = 0.110 [/tex]
[tex]Q = 10(0.110)(2840 - 2640) + (1450)(10)[/tex]
[tex]Q = 16300 k Cal[/tex]
