Question

A hot lump of 40.2g of aluminum at an initial temperature of 68.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

in the given problem we have to find out the final temperature of aluminum and water.

In this problem, as according to thermodynamics , the amount of heat given by aluminum should be equal to the amount of heat absorbed by water.

therefore, we first calculate the amount of heat , released by aluminum,

let it be Q1,

Then, Q1 = m*s*∆T

Where m is the mass of aluminum, s is specific heat of aluminum and ∆T is the temperature difference,

If Tf be the temperature difference, then , ∆T = 68.7-tf.

Now , Q1 = 40.2*0.903*(68.7-Tf).

= 36.3(68.7-Tf)

Q1 =2493.85 - 36.3 Tf

Now , we shall calculate the amount of heat absorbed by water,let it be Q2.

Q2= m.s.∆T

Since specific heat of water = 4.18 J/g°c

= 50*4.18*(Tf-25)

= 209 Tf - 5225

Since ,

Heat released by aluminum = Heat absorbed by water.

2493.85 - 36.3 Tf = 209 Tf - 5225

7718.85 = 245.3 Tf

Tf = 7718.85 /245.3

Tf = 31.46°C

therefore the final temperature = 31.46°C

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