How much heat is required to raise the temperature of 20.0g of H2O from 18.0 Celcius to 28.0 Celcius?
2 answers
heat=mass*specificheatcontent*deltaTemp
We have,
E = (mass)(specific heat capacity)(delta Temp)
mass = 20.0/1000 = 0.02 kg
=> E = (0.0200kg)(4200J/kg C)(28.0C-18.0C)
= 840 J
E = (mass)(specific heat capacity)(delta Temp)
mass = 20.0/1000 = 0.02 kg
=> E = (0.0200kg)(4200J/kg C)(28.0C-18.0C)
= 840 J