How many joules of heat are given off when 5.00g of water cool from 348.0K to 298.0K?

Q =  -1045 J

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 348.0 K

Final temperature = 298.0 K

Heat given off = ?

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.K

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 = 298.0 K - 348.0 K

ΔT = - 50 K

Q = 5.0 g ×4.18 J/g.K× - 50 K

Q =  -1045 J

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.