When 3.0 kg of water is warmed from 10.0°C to 80.0°C, how much heat energy is needed in kJ?

Question

When 3.0 kg of water is warmed from 10.0°C to 80.0°C, how much heat energy is needed in kJ?

Answer 1

Use the equation q = mC∆T

q = heat = ?

m = mass = 3.0 kg

C = specific heat of water = 4.184 kJ/kg/degree

∆T = change in temperature = 80.0º - 10.0º = 70ºC

q = mC∆T

q = (3.0 kg)(4.184 kJ/kg/ºC)(70ºC)

q = 878.6 kJ = 880 kJ (to 2 significant figures)

Answer 2

First, no phase change, so we just have to calculate

the amount of heat involved by the heat equation.

Q = mC_pΔT -------------------------------------------------------------where ΔT = T_f - T_i

Water is the substance used to define heat capacity which was set to 1 cal/g/°C.

Putting everything into the equation, we get:

Q = 3.0 Kg x 1000 g/Kg x 1 cal/g/°C x (80.0 - 10.0) °C --------units and significant figures are important here.

= 3.0 Kg x 1000 g/Kg x 1 cal/g/°C x (80.0 - 10.0) °C -----------only cal left.

= 210,000 cal

= 210 kcal 2 s.f.

Now convert to kJ

1 cal = 4.184 J

Q = 210 kcal x 4.184 kJ/kcal

= 878.64 kJ

= 880 kJ // answer --------------------------------------------------------2 s.f.