Asked by Reese

A 4.782 kg sample of water absorbs a certain quantity of heat. The temperature increases from 17.54 ºC to 26.34 ºC. Calculate how much heat the water absorbed. Specific heat constant for water is 4.184 J/g- ºC
Answered by DrBob222
q = heat absorbed = mass H2O x specific heat H2O x (Ttinal-Tinitial)
Substitute the numbers and solve. Post your work if you get stuck.
Answered by Reese
@dr bob

Sorry.

Q = m c delta-T
= (4.782 kg)*(4.184 J/gC)*(8.80 C) = around 180 kJ
Answered by DrBob222
not quite it. Notice that the mass you substituted is in kg but the secific heat is in units of J/GRAM*C. The easy to correct that is to change kg to grams, then recalculate.
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