How much heat is absorbed by 100.0 g of water warming from 20.0 °C to 23.5 °C? Use c = 4.184 J/(g*°C).
Solution
Problem
How much heat is absorbed by 100.0 g of water warming from 20.0 °C to 23.5 °C? Use c = 4.184 J/(g*°C).
Use q = mcΔT.
ΔT = 23.5 °C - 20.0 °C = 3.5 °C
q = (100.0 g)(4.184 J/(g*°C))(3.5 °C)
q = 1464 J
1464 J = 1.46 kJ.
Result
1.46 kJ