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

h let me think

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