How much thermal energy must be removed from 3.75 kg of water to freeze it
at its melting point? Use Q = mass x latent heat of fusion

D.1,250kj

Step-by-step explanation:

Hi, to answer this question we simply have to substitute mass= 3.75 and latent heat of fusion = 333 kj/kg in the equation given:

Q = mass x latent heat of fusion

Q = 3.75 kg x 333 kj/kg

Solving:

Q = 1,248 kj = 1,250kj (rounded)

The correct option is D.1, 250kj

Feel free to ask for more if needed or if you did not understand something.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

1. 8

2. 9

3. 20

4. 3

5. 17

6. 2.2

7. 11.25

8. 1

9. 14

10. 2

step-by-step explanation:

thats the first half