A solar heating specialist is considering paraffin as a possible solar heat collector. How many kilograms of paraffin would be needed to collect as much energy as 4.73 x 10^3 kg of water? The specific heat of paraffin is 2.90J/gxC. Hint: Assume the same temperature change.

Answer 1

You know by reasoning that it will be 4.18/2.90 = 1.44 x mass H2O
I would do this.
q for paraffin = mass P x specific heat x delta T.
q for H2O = 4730 x specific heat x delta T.
Set them equal.
mass P x 2.90 x delta T = 4730 x 4.18 x delta T.
mass paraffin = 4730 x 4.18/2.90 = ? and convert to kg and round.

Answer 2

The specific heat of water is 4.18*10^3 J/g*C.
The product of mass and specific heat would be the same for water and paraffin, for constant deltaT and energy content

The amount of paraffin needed to store the same heat as that amount of water would be higher by a factor
4.18/2.90 = 1.441

This assumes that the paraffin does not melt. A phase change would help store energy.