Respuesta :

Answer:

Specific heat of water = 33.89 KJ

Explanation:

Given:

mass of water = 81 gram

Initial temperature = 0°C

Final temperature = 100°C

Specific heat of water = 4.184

Find:

Required heat Q

Computation:

Q = Mass x Specific heat of water x (Final temperature - Initial temperature)

Q = (81)(4.184)(100-0)

Q = 33,890.4

Specific heat of water = 33.89 KJ

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