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Zielflug [23.3K]
3 years ago
7

Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls t

o water, what can be said about the final temperatures of the two balls? (Heat lost = mCpΔT, where m = mass of the object, Cp = specific heat capacity of the material, and ΔT = change in temperature).
Chemistry
1 answer:
Kobotan [32]3 years ago
5 0

Now we know that

Q = mc∆T

Where Q is y energy measured in Joules.

m is the mass measured in grams

c is the specific heat of the substance measured in joule per gram degree Celsius.

∆T is the change in temperature measured in degree Celsius.



Let Q1 be the specific heat of the lighter ball.

c1 be the specific heat of the lighter ball.

m1 be the mass of the lighter ball.

∆T1 be the change in the of the lighter ball.


Let Q2 be the specific heat of the heavier ball.

c2 be the specific heat of the heavier ball.

m2 be the mass of the heavierr ball.

∆T2 be the change in the of the heavier ball.


It has been given that the heat lost, that is Q is the same for both the balls of different mass.Which implies Q1= Q2

Specific heat(c) is the same for both the balls since both are made up of iron. c1=c2


Now heat lost by the lighter ball = heat lost by the heavier ball.

Q1= Q2

m1c1∆T1= m2c2∆T2

Since c1=c2

We get

m1/m2= ∆T2/∆T1

Thus we can say since m2>m1,∆T1> ∆T2.

Now initial temperature of both the balls are 100 degree Celsius.

∆T1 = Final temperature(T1 )-100.

∆T2= Final temperature ( T2)-100

Now since the ∆T1> ∆T2 as arrived from the above equation we can conclude that the final temperature of the ball 1 is greater than that of the ball 2. Since the ball 1 as per our assumption is the lighter ball,the final temperature of the ball which has lighter mass is greater than that of the one having a greater mass.

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Answer:

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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).

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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

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∴ 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.

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