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Need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow!!

Need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow-example-1
User Absessive
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1 Answer

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Answer:

Q14: 17,140 g = 17.14 kg.

Q16: 504 J.

Step-by-step explanation:

Q14:

  • To solve this problem, we can use the relation:

Q = m.c.ΔT,

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.

Q16:

  • To solve this problem, we can use the relation:

Q = m.c.ΔT,

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.

User Pomaxa
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