Question

If you add 700 kJ of heat to 700 g of water originally at 70.0°C, how much water is left in the container? The latent heat of vaporization of water is 22.6 × 105 J/kg, and its specific heat capacity is .

Answer 1

700,000=mx22.6x10^5 ... 7=mx22.6 ... m=7/22.6 ... ~ 7/21 ~0.33kg  evaporated. ... 330g gone, 640 left

Answer 2

We have that the amount of water left in the container is

m'=0.4kg

From the question we are told

If you add 700 kJ of heat to 700 g of water originally at 70.0°C

The latent heat of vaporization of water is 22.6 x 105 J/kg

Generally the equation for 100C Heat  is mathematically given as

$Q = m C(T2-T1)\\\\Q = 0.700 4186(100-70)\\\\Q=8.8 x 10^4J$

Generally the equation for the evaporation water Heat  is mathematically given as

$Q'=Q1-Q\\\\Q'=7 x 10^5-8.79x 10^4\\\\\Q'61.21 x 10^4J$

Hence with

$Q'=Lm\\\\m=\frac{61.21 x 10^4}{23*10^5}\\\\m=0.3kg$

Therefore

The total Remaining within the contain will be

$m'=0.700-0.3$

m'=0.4kg

Therefore

The amount of water left in the container is

m'=0.4kg

For more information on this visit