Question

How many joules are required to heat 250.00 grams of water at 15.5 degrees Celsius to its boiling point, 100.0 degrees Celsius

Answer 1

Q = 88725 J

Step-by-step explanation:

Q = mcΔT

where,

c = 4.2 J/g°C

m = 250 g

ΔT = Tfinal − Tinitial = 100ºC − 15.5ºC = 84.5ºC

Q = 250 x 4.2 x 84.5

Q = 88725