If you start with 163 g of water at 29◦C, how much heat must you add to convert all the liquid into vapor at 100◦C? Assume no heat is lost to the surroundings.
Answer in units of kJ.

48.37514 kj

Explanation:

Given data:

Mass of water = 163 g

Initial temperature = 29°C

Final temperature = 100°C

Heat added = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT =  100°C - 29°C

ΔT =  71°C

Q = 163 g × 4.18 j/g.°C × 71°C

Q = 48375.14 j

Joule to Kj conversion:

48375.14 /1000 = 48.37514 kj

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

idk what the question is about