Answer:
39.57 °C
Explanation:
Applying,
Heat gain = Heat lost.
cm'(t₃-t₁) = cm(t₂-t₃)................. Equation 1
Where c = specific heat capacity of water, m = mass of the hotter water, m' = mass of the colder water, t₁ = initial temperature of the colder water, t₂ = initial temperature of the hotter water, t₃ = Final Temperature.
Equation 1 above can futher be simplified to
m'(t₃-t₁) = m(t₂-t₃)................. Equation 2
From the question,
Given: m' = 155 g, m' = 75 g, t₁ = 20 °C, t₂ = 80 °C
Substitute these values into equation 2
155(t₃-20) = 75(80-t₃)
Solve for t₃
155t₃- 3100 = 6000-75t₃
155t₃+75t₃ = 6000+3100
230t₃ = 9100
t₃ = 9100/230
t₃ = 39.57 °C
Answer:
C) Remove some of the water. Cover the rocks with an absorbent like kitty litter to remove the oil. Remove the dirty kitty litter. Add the water back.
Explanation:
Oil and water do not mix. The oil will cover the water's surface and anything in the water. So the best idea would be to absorb the oil from the system.
<span>small K = positive ∆G = negative E0cell</span>
The moon phases maybe? Can I see the passage possibly?
This would be the downward movement of water. It begins flowing and continues in a downward movement moving throughout,
