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nydimaria [60]
2 years ago
8

what amount of heat is removed to lower the temperature of 80 grams of water from 75°C to 45°C? the specific heat of liquid wate

r is 4.18J/g°C
Physics
2 answers:
Slav-nsk [51]2 years ago
3 0

Answer: option A.

Explanation:

Q= heat removed from the water

m= mass of the water = 80 grams

c = heat capacity of water= 4.18 J/g°C      

 

Q = -10,032 joules  -10,000 Joules

Negative sign indicates that heat was removed from the water

MaRussiya [10]2 years ago
3 0

Answer:

10032 J    must be removed

Explanation:

80 gm  *  (75-45 C) * 4.18 J/g-C  = 10032 J

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Read 2 more answers
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