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 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.)

m_m = 0.500 kg

T_im = 100⁰C

T_f = 22.1⁰C

m_w = 0.400 kg

T_iw = 20⁰C

T_fw = 22.1⁰C

c_w = 4186 J/kg ·⁰C

c_m = ?

          Heat lost + Heat gained = 0            

Q = mcΔT

m_mc_mΔT_m + m_wc_wΔT_w = 0

Plug in the numbers:

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

Calculate the change in temperautres:

0.0500 kg(c_m)(-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.)

m_I = 0.022 kg

T_iI = 0⁰C

L_f = 3.33 x 10⁵ J/kg

m_w = 0.450 kg

T_iw = 80⁰C

c_w = 4186 J/kg·⁰C

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_IL_f + m_Ic_wΔT_I + m_wc_wΔT_w = 0

Plug in the numbers:

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

Multiply:

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

Combine like terms:

1975.792 J/⁰C(T_f) -143370 J = 0

Add 143370 J to both sides:

1975.792 J/⁰C(T_f) = 143370 J

Divide both sides by 1975.792 J/⁰C

T_f = 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.)

m_c = 0.312 kg

T_ic = 99⁰C

T_fc = 25⁰C

c_c = ?

m_w = 0.507 kg

T_iw = 21⁰C

T_fw = 25⁰C

c_w = 4186 J/kg·⁰C

m_a = 0.105 kg

T_ia = 21⁰C

T_fa = 25⁰C

c_a = 889 J/kg·⁰C                                       

Heat lost + Heat gained = 0

m_cc_cΔT_c + m_wc_wΔT_w + m_ac_aΔT_a = 0

Plug in the numbers:

0.312 kg(c_c)(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(c_c)(-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(c_c) + 8489.208 J + 373.38 J = 0

Combine like terms:

-23.088 kg·⁰C(c_c) + 8862.588 J = 0

Add 23.088 kg·⁰C(c_c) to both sides:

8862.588 J = 23.088 kg·⁰C(c_c)

Divide both sides by 23.088 kg·⁰C:

c_c = 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.)

m_L = 0.853 kg

c_L = 128 J/kg·⁰C

T_iL = ?

T_f = 5⁰C

m_I = 0.015 kg

c_I = 209 J/kg·⁰C

T_iI = -5⁰C

T_fI = 0⁰C

L_f = 3.33 x 10⁵ J/kg

c_w = 4186 J/kg·⁰C

T_iw = 0⁰C

T_fw = 5⁰C       

Heat lost + Heat gained = 0 

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

Then the ice has to melt     Q = m_ILf

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

m_Lc_LΔT_L +  m_Ic_IΔT_I + m_ILf + m_Ic_wΔT_w  = 0

Plug in the numbers:

0.853 kg(128 J/kg·⁰C)(5⁰C - T_i) + 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(T_i) + 0 J + 15.675 J + 4995 J + 313.95 J - 0 J = 0

Combine like terms:

5870.545J -109.184 J/⁰C(T_i) = 0

Add 109.184 J/⁰C(T_i) to both sides:

Divide both sides by 109.184 J/⁰C

T_i = 53.8⁰C