What is the heat in Joules required to raise the temperature of 100g H20 from 288K to 308K?

Answer 1

to get the heat released or absorbed,
Q = mc(T2-T1)
where
m = mass of substance (units in g)
c = specific heat capacity (units in J/g-K)
T2 = final temperature
T1 = initial temperature
**note: if Q is (-), heat is released and if (+), heat is absorbed

for water c = 4.184 J/g-K
substituting,
Q = 100*4.184*(308-288)
Q = 8368 J

hope this helps~ :)

Answer 2

8368J