Specific heat probelem q = mSΔT?

Question

Help, I need to know

How many KJ of heat are absorbed when 1,000 grams of water is heated from

18 0 C to 85 0 C??

This is just a practice probelem, already have the answer, but could you write it out in detail?

What would specific heat be on this probelem? (S)

q = mSΔT

Thanks so much~! :3

Answer 1

q = ?

m = 1, 000 g

c = 4.184 g/j/c

Tf = 85 - 18 = 67 C

q = mct

q = (1000 g) (4.184 C) (67C)

q = 280328 J

q = 280.33 KJ of Heat.

Answer 2

If q is the quantity of heat absorbed, it is equal to the product of

the mass times specific heat (also called heat capacity) times the

change in temperature.  Here the change in temp. is given as

85 - 18 or 67°, the mass of water is 1000 grams and the heat

capacity or specific heat of water is 1 calorie per gram per degree.

To find it in joules, 1 calorie = 4.184 joules, so for 1000 grams of

water, the heat capacity is 4184 joules per degree.  The change in

temperature is 67° so the total heat absorbed is 4184 times 67,

or 280,328 joules, or 280.328 kilojoules.

Hope this answers your question.