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,

$\begin{gathered} Q=(1.70\text{ kg})\times(4.186\text{ J/g}^{\circ}C)\times(100^{\circ}C-23^{\circ}C) \\ =(1.70\times10^3\text{ g})\times(4.186\text{ J/g}^{\circ}C)\times(100^{\circ}C-23^{\circ}C) \\ =547947.4\text{ J} \end{gathered}$

Converting Joules to cal;

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

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