Simplified Heat Problems

by

Marilynn Stone

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         Most text books are very confusing when it comes to solving heat problems.  I have discovered an easier way to do these problems.

Just use:    Heat lost + Heat gained = 0

If you use this simple equation, ΔT is always Tf - Ti, and you don't have to figure out which substances are losing heat and which are gaining heat, just plug in all the numbers and solve.

 

Examples:

1.  A metal sample with a mass of 0.0500 kg at 1000C is placed into 0.400 kg a water at 20.00C.  If the final temperature is 22.10C,  what is the specific heat of the metal sample?  (Assume that there is no heat lost to the surroundings.)

mm = 0.500 kg

Tim = 1000C

Tf = 22.10C

mw = 0.400 kg

Tiw = 200C

Tfw = 22.10C

cw = 4186 J/kg ·0C

cm = ?

          Heat lost + Heat gained = 0            

Q = mcΔT

mmcmΔTm + mwcwΔTw = 0

Plug in the numbers:

0.0500 kg(cm)(22.10C - 1000C) + 0.400 kg(4186 J/kg ·0C)(22.10C - 200C) = 0

Calculate the change in temperautres:

0.0500 kg(cm)(-77.90C) + 0.400 kg(4186 J/kg ·0C)(2.10C) = 0

Multiply:

     -3.895 kg·0C (c) + 3516.24 J = 0 

Add 3.895 kg·0C (c) to both sides:

3516.24 J = 3.895 kg·0C(c)

Divide both sides by 3.895 kg·0C:

c = 902.76 J/kg·0C

 

2.  If 0.022 kg of ice at 00C is added to 0.450 kg of water at 800C, what is the final temperature. (Assume that there is no heat lost to the surroundings.)

mI = 0.022 kg

TiI = 00C

Lf = 3.33 x 105 J/kg

mw = 0.450 kg

Tiw = 800C

cw = 4186 J/kg·0C

Tf = ?

Heat lost + heat gained = 0

Q = mLf          Q = mcΔT

The ice has to melt first before the final temperature can be found.

mILf + mIcwΔTI + mwcwΔTw = 0

Plug in the numbers:

0.022 kg(3.33 x 105 J/kg) + 0.022 kg(4186 J/kg·0C)(Tf - 00C) + 0.450 kg(4186 J/kg·0C)(Tf - 800C) = 0

Multiply:

7326 J + 92.092 J/0C(Tf) - 00C + 1883.70 J/0C(Tf) - 150696 J = 0

Combine like terms:

1975.792 J/0C(Tf) -143370 J = 0

Add 143370 J to both sides:

1975.792 J/0C(Tf) = 143370 J

Divide both sides by 1975.792 J/0C

Tf = 72.560C

 

3.  A copper sample with a mass of 0.312 kg is at 990C.  The sample is placed into 0.507 kg of water at 210C which is in an aluminum calorimeter  which has a mass of 0.105 kg.  If the final temperature is 250C, calculate the specific heat of copper.  (Assume that there is no heat lost to the surroundings.)

mc = 0.312 kg

Tic = 990C

Tfc = 250C

cc = ?

mw = 0.507 kg

Tiw = 210C

Tfw = 250C

cw = 4186 J/kg·0C

ma = 0.105 kg

Tia = 210C

Tfa = 250C

ca = 889 J/kg·0C                                       

Heat lost + Heat gained = 0

mcccΔTc + mwcwΔTw + macaΔTa = 0

Plug in the numbers:

0.312 kg(cc)(250C - 990C) + 0.507 kg(4186 J/kg·0C)(250C - 210C) + 0.105 kg(889 J/kg·0C)(250C - 210C) = 0

Calculate the change in temperatures:

0.312 kg(cc)(-740C) + 0.507 kg(4186 J/kg·0C)(4.00C) + 0.105 kg(889 J/kg·0C)(4.00C) = 0

Multiply:

-23.088 kg·0C(cc) + 8489.208 J + 373.38 J = 0

Combine like terms:

-23.088 kg·0C(cc) + 8862.588 J = 0

Add 23.088 kg·0C(cc) to both sides:

8862.588 J = 23.088 kg·0C(cc)

Divide both sides by 23.088 kg·0C:

cc = 384 J/kg·0C

( the accepted value for copper is 387 J/kg·0C)

 

4.  To what temperature must a 0.853 kg block of lead be heated in order to completely melt 0.015 kg of ice which is initially at -50C.  The final temperature of the melted ice is +50C. (Assume that there is no heat lost to the surroundings.)

mL = 0.853 kg

cL = 128 J/kg·0C

TiL = ?

Tf = 50C

mI = 0.015 kg

cI = 209 J/kg·0C

TiI = -50C

TfI = 00C

Lf = 3.33 x 105 J/kg

cw = 4186 J/kg·0C

Tiw = 00C

Tfw = 50C       

Heat lost + Heat gained = 0 

First the ice must heat up to 00C    Q = mIcIΔTI

Then the ice has to melt     Q = mILf

Next the melted ice has to warm up to 50C    Q = mIcwΔTw

mLcLΔTLmIcIΔTI + mILf + mIcwΔTw  = 0

Plug in the numbers:

0.853 kg(128 J/kg·0C)(50C - Ti) + 0.015kg(209 J/kg·0C )[00C - (-50C)] + 0.015kg(3.33 x 105J/kg) + 0.015 kg(4186J/kg·0C )(50C - 00C) = 0

Multiply:

545.92 J - 109.184 J/0C(Ti) + 0 J + 15.675 J + 4995 J + 313.95 J - 0 J = 0

Combine like terms:

5870.545J -109.184 J/0C(Ti) = 0

Add 109.184 J/0C(Ti) to both sides:

Divide both sides by 109.184 J/0C

Ti = 53.80C

 

 

 

 

 

 

                                                                                                                                                             

                                            

 

 

                                                                                                                                                                   

                                               

           

                                                                                                                                                                     

                                                 


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