Suppose you want to heat 40kg of water by 20 deg C How many Joules of heat are required?

Answer 1

You would use the equation ΔQ = mcΔT, where ΔQ means change in energy (or energy required in this case), m is mass of water, c is specific heat capacity, and ΔT change in temperature. Water's specific heat capacity (energy needed to raise 1kg by 1 degree K or C) is 4200J Kg-1 K-1, so, if we put your numbers in: Q = 40kg x 4200J Kg-1 K-1 x 20°C (20 degress kelvin)

Q = 3360000J, or, 3360KJ, or 3.36MJ of energy. This is assuming that the temperature change does not get the water to boil, as then you have to factor in the latent heat of vapourisation...

Answer 2

The formula that relates energy and change in temperature for a substance is q=mcT where q is energy in Joules, m is mass in grams, c is the specific heat of the substance in J/gK, and T is the change in temperature in degrees Kelvin (or celcius).

You have 40kg of water which is 40,000 grams. = m

The specific heat of water is 4.186 J/gK = c

The change in temperature is 20 C or 20 K = T

q = (40000)(4.186)(20) = 3348800 Joules
You would use the equation ΔQ = mcΔT , where ΔQ means change in energy (or energy required in this case), m is mass of water, c is specific heat capacity, and ΔT change in temperature. Water's specific heat capacity (energy needed to raise 1kg by 1 degree K or C) is 4200J Kg-1 K-1, so, if we put your numbers in: Q = 40kg x 4200J Kg-1 K-1 x 20°C (20 degress kelvin)

Q = 3360000J, or, 3360KJ, or 3.36MJ of energy. This is assuming that the temperature change does not get the water to boil, as then you have to factor in the latent heat of vapourisation...