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
Answer:
two may be the answer (2)
