Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
Answer:
Total worth of gold in the ocean = $5,840,000,000,000,000
Explanation:
As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.
Therefore, In 1 L of ocean water there will be,
(4.0 x 10^-10)/0.0021
= 1.9045 x 10^-7 g of gold per Liter of ocean water.
So in 1.5 x 10^-21 L of ocean water, there will be
(1.9045 x 10^-7) * (1.5 x 10^-21)
= 2.857 x 10^14 g of gold in the ocean.
1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is
20.44 * (2.857 x 10^14)
= $5,840,000,000,000,000
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