q = heat absorbed = mass H2O x specific heat H2O x (Ttinal-Tinitial)
Substitute the numbers and solve. Post your work if you get stuck.
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
3 answers
@dr bob
Sorry.
Q = m c delta-T
= (4.782 kg)*(4.184 J/gC)*(8.80 C) = around 180 kJ
Sorry.
Q = m c delta-T
= (4.782 kg)*(4.184 J/gC)*(8.80 C) = around 180 kJ
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.
1 answer