How much heat is required to heat 1.8 kg of water from 8.4oC to 17.3oC? The heat capacity of water is 4.1813 J/g*oC

Q = 66984.43 J

Explanation:

Given data:

Mass of water = 1.8 Kg (1.8Kg×1000 g/1 Kg =1800 g)

Initial temperature = 8.4°C

Final temperature = 17.3°C

Specific heat capacity of water = 4.1813 J/g.°C

Heat required = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 17.3°C - 8.4°C

ΔT = 8.9°C

Q = 1800 g ×  4.1813 J/g.°C ×8.9°C

Q = 66984.43 J

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

explanation:

a