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

Answer:The correct answer option A.

Explanation:

$Q= m\times c\times \Delta T$

Q= heat removed from the water

m= mass of the water = 80 grams

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

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

$Q=80g\times 4.18J/g^oC\times(-30^oC)$

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.

Answer 2

Answer:

The correct answer is -10000J. The option A

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