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30 votes
30 votes
I NEED HELP!!!

How much heat (kJ) is needed to raise the temperature of 100.0 grams of water from 290.0 K to 325.0 K?
(Specific Heat of Water = 4.18 J/g °C)
• 10450
• 146
• 14630
• 10.5
• 14.6

User Vidang
by
3.0k points

1 Answer

13 votes
13 votes

Answer:

Step-by-step explanation:

Formula

Heat = m * c * delta t

t = temperature in centigrade.

The first thing you have to do is convert kelvin degrees to centigrade. The conversion factor is - 273. The formula is degrees centigrade = degrees kelvin - 273. It is easier to understand with a couple of examples.

  • 290o K = 290 - 273 = 17
  • 325. K = 325 - 273 = 52

Solution

Heat = 100.0 grams * 4.18 J/g*C * (52 - 17)

Heat = 100 * 4.18 * 35

Heat = 14630 Joules

But you want kj

Heat = 14630 / 100 = 14.63

Answer

Heat = 14.6 kj

User Shankshera
by
3.2k points