how many joules are needed to warm 25.5 grams of water from 14.0°C to 22.5°C? the heat capacity of liquid water is 4.184J/g°C

Q = 906.882 J

Explanation:

Given data:

Mass of water = 25.5 g

Initial temperature = 14.0 °C

Final temperature = 22.5 °C

Specific heat capacity of water = 4.184 J/g.°C

Solution:

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 = 22.5 °C - 14.0 °C

ΔT = 8.5 °C

Q = 25.5 g × 4.184 J/g.°C ×  8.5 °C

Q = 906.882 J