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Just use: **
Heat lost + Heat gained = 0**

If you use this simple equation, ΔT is always T_{f}
- T_{i}, 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 100^{0}C
is placed into 0.400 kg a water at 20.0^{0}C. If the
final temperature is 22.1^{0}C, what is the specific heat
of the metal sample? (Assume that there is no heat lost to the
surroundings.)

m_{m} = 0.500 kg

T_{im} = 100^{0}C

T_{f} = 22.1^{0}C

m_{w} = 0.400 kg

T_{iw} = 20^{0}C

T_{fw} = 22.1^{0}C

c_{w} = 4186 J/kg^{
}·^{0}C

c_{m} = ?

Heat lost + Heat gained = 0

Q = mcΔT

m_{m}c_{m}ΔT_{m} + m_{w}c_{w}ΔT_{w}
= 0

Plug in the numbers:

0.0500 kg(c_{m})(22.1^{0}C - 100^{0}C)
+ 0.400 kg(4186 J/kg^{ }·^{0}C^{)(}22.1^{0}C
- 20^{0}C) = 0

Calculate the change in temperautres:

0.0500 kg(c_{m})(-77.9^{0}C) + 0.400
kg(4186 J/kg^{ }·^{0}C)(2.1^{0}C)
= 0

Multiply:

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

Add 3.895 kg·^{0}C^{
}(c) to both sides:

3516.24 J = 3.895 kg·^{0}C(c)

Divide both sides by 3.895 kg·^{0}C:

c = 902.76 J/kg·^{0}C

2. If 0.022 kg of
ice at 0^{0}C is added to 0.450 kg of water at 80^{0}C,
what is the final temperature. (Assume that there is no heat lost to
the surroundings.)

m_{I} = 0.022 kg

T_{iI} = 0^{0}C

L_{f} = 3.33 x 10^{5 }J/kg

m_{w} = 0.450 kg

T_{iw} = 80^{0}C

c_{w} = 4186 J/kg·^{0}C

T_{f} = ?

Heat lost + heat gained = 0

Q = mL_{f}
Q = mcΔT

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

m_{I}L_{f} + m_{I}c_{w}ΔT_{I}
+ m_{w}c_{w}ΔT_{w} = 0

Plug in the numbers:

0.022 kg(3.33 x 10^{5 }J/kg) + 0.022 kg(4186
J/kg·^{0}C)(T_{f} -
0^{0}C) + 0.450 kg(4186 J/kg·^{0}C)(T_{f}
- 80^{0}C) = 0

Multiply:

7326 J + 92.092 J/^{0}C(T_{f})
- 0^{0}C + 1883.70 J/^{0}C(T_{f}) - 150696 J = 0

Combine like terms:

1975.792 J/^{0}C(T_{f})
-143370 J = 0

Add 143370 J to both sides:

1975.792 J/^{0}C(T_{f})
= 143370 J

Divide both sides by
1975.792 J/^{0}C

T_{f} = 72.56^{0}C

3. A copper sample
with a mass of 0.312 kg is at 99^{0}C. The sample is
placed into 0.507 kg of water at 21^{0}C which is in an
aluminum calorimeter which has a mass of 0.105 kg. If the
final temperature is 25^{0}C, calculate the specific heat of
copper. (Assume that there is no heat lost to the surroundings.)

m_{c} = 0.312 kg

T_{ic} = 99^{0}C

T_{fc} = 25^{0}C

c_{c} = ?

m_{w} = 0.507 kg

T_{iw} = 21^{0}C

T_{fw} = 25^{0}C

c_{w} = 4186 J/kg·^{0}C

m_{a} = 0.105 kg

T_{ia} = 21^{0}C

T_{fa} = 25^{0}C

c_{a }= 889 J/kg·^{0}C

Heat lost + Heat gained = 0

m_{c}c_{c}ΔT_{c}
+ m_{w}c_{w}ΔT_{w} + m_{a}c_{a}ΔT_{a}
= 0

Plug in the numbers:

0.312 kg(c_{c})(25^{0}C - 99^{0}C)
+ 0.507 kg(4186 J/kg·^{0}C)(25^{0}C - 21^{0}C)
+ 0.105 kg(889 J/kg·^{0}C)(25^{0}C
- 21^{0}C) = 0

Calculate the change in temperatures:

0.312 kg(c_{c})(-74^{0}C) + 0.507 kg(4186
J/kg·^{0}C)(4.0^{0}C)
+ 0.105 kg(889 J/kg·^{0}C)(4.0^{0}C)
= 0

Multiply:

-23.088 kg·^{0}C(c_{c}) + 8489.208
J + 373.38 J = 0

Combine like terms:

-23.088 kg·^{0}C(c_{c}) + 8862.588
J = 0

Add 23.088 kg·^{0}C(c_{c}) to both
sides:

8862.588 J = 23.088 kg·^{0}C(c_{c})

Divide both sides by
23.088
kg·^{0}C:

c_{c} = 384 J/kg·^{0}C

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

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 -5^{0}C.
The final temperature of the melted ice is +5^{0}C. (Assume
that there is no heat lost to the surroundings.)

m_{L} = 0.853 kg

c_{L} = 128 J/kg·^{0}C

T_{iL} = ?

T_{f} = 5^{0}C

m_{I} = 0.015 kg

c_{I} = 209 J/kg·^{0}C

T_{iI} = -5^{0}C

T_{fI} = 0^{0}C

L_{f} = 3.33 x 10^{5
}J/kg

c_{w} = 4186 J/kg·^{0}C

T_{iw} = 0^{0}C

T_{fw} = 5^{0}C

Heat lost + Heat gained = 0

First the ice must heat
up to 0^{0}C Q = m_{I}c_{I}ΔT_{I}

Then the ice has to melt Q = m_{I}Lf

Next the melted ice has to warm up to 5^{0}C
Q = m_{I}c_{w}ΔT_{w}

m_{L}c_{L}ΔT_{L} +
m_{I}c_{I}ΔT_{I}
+ m_{I}Lf + m_{I}c_{w}ΔT_{w} = 0

Plug in the numbers:

0.853 kg(128 J/kg·^{0}C)(5^{0}C
- T_{i}) + 0.015kg(209 J/kg·^{0}C
)[0^{0}C - (-5^{0}C)] + 0.015kg(3.33 x 10^{5}J/kg)
+ 0.015 kg(4186J/kg·^{0}C
)(5^{0}C - 0^{0}C) = 0

Multiply:

545.92 J - 109.184 J/^{0}C(T_{i})
+ 0 J + 15.675 J + 4995 J + 313.95 J - 0 J = 0

Combine like terms:

5870.545J -109.184 J/^{0}C(T_{i})
= 0

Add 109.184 J/^{0}C(T_{i})
to both sides:

Divide both sides by
109.184 J/^{0}C

T_{i} = 53.8^{0}C

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