Question

What amount of heat is removed to lower the temperature of 80 grams of water from 75º C to 45º C? The specific heat of liquid water is 4.18 J/gº C.

A. -10,000 J
B. 10,000 J
C. -8,000 J
D. 8,000 J

Answer 1

The correct answer is -10000J. The option A

Step-by-step explanation:

First we have to see the data we have

m=80g

c=4.18J/g*°C

T2=45°C

T1=75°C

ΔT=T2-T1=(45-75)°C=-30°C

The formula of heat is

Q=m*c*ΔT

Q=80g* 4.18J/g*°C*-30°C

Q=-10032J

The correct answer is -10000J

Answer 2

Answer:The correct answer option A.

Step-by-step explanation:

`Q= m* c* \Delta T`

Q= heat removed from the water

m= mass of the water = 80 grams

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

`\Delta T={\text{Change in temperature}}=45^oC-75^oC=-30^oC`

`Q=80g* 4.18J/g^oC*(-30^oC)`

Q = -10,032 joules
`\approx` -10,000 Joules

Negative sign indicates that heat was removed from the water.

Heat removed from the water of 80 gram of water is 10,032 Joules.Hence, the correct answer option A.