The specific heat of water is 4.184 J/g˚C How much heat energy is absorbed when 88.0 g of water is heated from 5.8 C to 35 C? (H
int: it is just like example#1). Don't forget units and round your answer to the second decimal place. *
1 answer:
Answer:
The heat absorbed is hence 10.751.21 J
Explanation:
The heat absorbed when 88g of water is heated from 5.8°C to 35 °C is;
Heat = m c ΔT
m = 88.0 g
c = specific heat of water = 4.184 J/g°C
ΔT = ( change in temperature) = ( 35 - 5.8)°C = 29.2°C
Equating these values into the formula, we obtain;
Heat = 88* 4.184 * 29.2
Heat = 10 751.2064 J
Heat = 10 751.21 J (2 d.p)
The heat absorbed is hence 10.751.21 J
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