Question

How much energy in Cal does it take to completely boil 1.70 kg of water at 23oC?

Answer 1

Given data:

Mass of water:

`m=1.70\text{ kg}`

Initial temperature of water:

`T_i=23^(\circ)C`

Energy required to boil water is given as,

`Q=mC_p(T_b-T_i)`

Here, Cp is the specific heat capactity of water, and Tb is the boiling temperature of water.

Substituting all known values,

`\begin{gathered} Q=(1.70\text{ kg})*(4.186\text{ J/g}^(\circ)C)*(100^(\circ)C-23^(\circ)C) \\ =(1.70*10^3\text{ g})*(4.186\text{ J/g}^(\circ)C)*(100^(\circ)C-23^(\circ)C) \\ =547947.4\text{ J} \end{gathered}`

Converting Joules to cal;

`\begin{gathered} Q=(547947.4\text{ J})*(\frac{0.239006\text{ cal}}{1\text{ J}}) \\ =130962.71\text{ cal} \end{gathered}`

Therefore, 130962.71 cal of energy required to completely boil 1.70 kg of water at 23°C.