Question

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 water is 4.18J/g°C

Answer 1

10032 J must be removed

Step-by-step explanation:

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

Answer 2

Answer: option A.

Step-by-step 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