Question

MULTIPLE CHOICE!!! YOU CAN ANSWER ANY ONE YOU WANT!!! PLEASE I NEED HELP FAST!

Calculate the total heat content of 10 kg of ice at -23°C. (Hint: A substance at absolute zero, -273o C, has 0 heat content.)
115 kcal
1300 kcal
1400 kcal
1500 kcal

How much heat is contained in 100 kg of water at 60.0 °C? (Hint: A substance at absolute zero, -273o C, has 0 heat content.)
30 kcal
2.10 x 104 kcal
2.40 x 104 kcal
2.77 x 104 kcal

Calculate the energy needed to vaporize 10.0 kg of iron (C = 0.110) initially at 2,640°C.
110
16,300

Answer 1

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

1.

Answer:

1300 k Cal

Step-by-step explanation:

Total heat contained by ice at -23 degree C is given by

`Q = ms\Delta T`

here we know that

`m = 10 kg`

s = 500 Cal/kg K

`\Delta T = (-23) - (-273)`

`\Delta T = 250`

now by above formula we have

`Q = 10(500)(250)`

`Q = 1300 kCal`

2

Answer:

`2.77 * 10^4 k Cal`

Step-by-step explanation:

Total heat contained by water at 60 degree C is given by

`Q = ms_i\Delta T + mL + ms_w\Delta T'`

here we know that

`m = 100 kg`

`s_w = 1000 Cal/kg K`

`s_i = 500 cal/kg K`

`\Delta T = 0 - (-273) = 273`

`\Delta T' = (60) - (0)`

`\Delta T' = 60`

now by above formula we have

`Q = 100(500)(273) + 100(80000) + 100(1000)(60)`

`Q = 2.77* 10^4 kCal`

3

Answer:

`16,300 k Cal`

Step-by-step explanation:

Energy required to vaporize the iron is given as

`Q = ms\Delta T + mL`

`m = 10 kg`

`C = 0.110`

`Q = 10(0.110)(2840 - 2640) + (1450)(10)`

`Q = 16300 k Cal`