How much energy was required to heat a 5.00g sample of liquid water at 25.0°C to 100°C? (Specific heat of water is 4.184J/g°C).

Q = 1569 j

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 25.0°C

Final temperature = 100°C

Cp = 4.184 j/g.°C

Energy needed = ?

Solution:

Specific heat capacity:

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

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 =  100°C - 25°C

ΔT =  75°C

Q = 5 g× 4.184 j/g.°C×75°C

Q = 1569 j