Answer:
The specific heat of water is 4.18 J/g C.
Explanation:
q
=
m
C
s
Δ
T
Never forget that!
2200
=
m
⋅
4.18
J
g
⋅
°
C
⋅
66
°
C
∴
m
≈
8.0
g
<u>Answer:</u>
<em>When we finish, the temperature would be 32.5℃</em>
<em></em>
<u>Explanation:</u>
Density of water = mass/volume
So,
Mass of water = Density × Volume


where
= Final T - Initial T
Q is the heat energy in calories
c is the specific heat capacity (for water 1.0 cal/(g℃))
m is the mass of water
plugging in the values


Final T = ∆T + Initial T
= 7.5℃ + 25℃ = 32.5℃ (Answer).
Answer:
B. The effects of warmed water on aquatic life
Explanation:
quizizz
Hope this helps! 29.221385