What is the final temperature of 10.0g of water initially at C after absorption of 100.J of heat?

Question

What is the final temperature of 10.0g of water initially at C after absorption of 100.J of heat?

The specific heat of water is 4.184 J/g C.

Answer 1

Here we go again...

q = mC∆T

q = heat = 100 J

m = mass = 10.0 g

C = specific heat = 4.184 J/g/degree

∆T = change in temperature = ?

Solve for ∆T

∆T = q/(m)(C) = 100 J/(10.0 g)(4.184 J/g/degree)

∆T = 2.39 degrees

Final temperature = 25º + 2.39 degrees = 27.39 degrees = 27 degrees C(2 sig. figs.)