A 5.00-g sample of aluminum pellets (specific heat capacity 0.89 J/g°C) and a 10.00-g sample of iron pellets (specific heat capacity 0.45 J/g°C) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Question

High School

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Answer :

Qwbox Qwbox · Answer 1 · 2023-09-18T05:38:06-04:00

The final temperature of the metal and water mixture when no heat loss to the surrounding is 23.7 degree Celsius.

The mass of the aluminium Pellet is 5 gram, specific heat of Aluminium is given to be 0.89 J/C-g

The mass of iron pellets is 10 gram and specific heat of iron is given to be 0.45 J/C-g.

They are heated to a temperature of 100° Celsius they are dropped into water of mass 97.3 grams which is at temperature of 22° Celsius.

As there is no heat loss,

The total heat eliminated by aluminium and will be equal to the heat gained by water,

So, we can write, heat released or gained,

Q = mC∆T

Where,

Q is the heat eliminated by aluminium and iron,

M is the mass of substance,

∆T is the change and temperature.

If we assume that T is the final temperature of the system when aluminium and iron are dropped in the water,

For iron,

Q = (5)(0.89)(100-T)

Q = (10)(0.45)(100-T)

For water,

Q = (97.3)(4.184)(T-22)

So,

Heat lost by aluminium and iron = heat gained by water,

(5)(0.89)(100-T) + (10)(0.45)(100-T) = (97.3)(4.184)(T-22)

T = 23.7° C.

The final temperature of the system is 23.7 degree Celsius

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