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=(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

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