Question

If we are tying to heat 90 g of water to a boil starting at 22 degrees C, how much energy in joules will we need?

The ‘c’ value for water is 4.18 J/gC

Q=mC(delta)T

Answer 1

29000J (2sf)

Step-by-step explanation:

Q = m•c•ΔT

= (90)(4.18)(78)

= 29343.6J

= 29000J (2sf)

NOTE: m=mass of water

c = specific heat capacity of water (4.18J/g/K)

ΔT = the change in temperature (100-22=78)