8. challenge a $4.00 \\times 10^2$-g sample of water at $15.0^circ$c is mixed with $4.00 \\times 10^2$ g of…

Explanation:

Step 1: Find equilibrium temp of two waters

Let ( T_1 ) be temp after mixing two waters.
Heat lost by hot water = Heat gained by cold water.
( m_1c\Delta T_1 = m_2c\Delta T_2 ) (c cancels, ( m_1 = m_2 = 4.00\times10^2,\text{g} ))
( 400(85 - T_1) = 400(T_1 - 15) )
( 85 - T_1 = T_1 - 15 )
( 2T_1 = 100 )
( T_1 = 50.0^\circ\text{C} )

Step 2: Mix methanol with water mixture

Let ( T_f ) be final temp.
Mass of water mixture: ( 400 + 400 = 800,\text{g} ), ( T_{water} = 50^\circ\text{C} )
Mass of methanol: ( 400,\text{g} ), ( T_{methanol} = 15^\circ\text{C} )
( c_{water} = 4.18,\text{J/g}^\circ\text{C} ), ( c_{methanol} = 2.51,\text{J/g}^\circ\text{C} )

Heat lost by water = Heat gained by methanol:
( 800 \times 4.18 \times (50 - T_f) = 400 \times 2.51 \times (T_f - 15) )

Simplify:
( 800 \times 4.18 \times 50 - 800 \times 4.18 T_f = 400 \times 2.51 T_f - 400 \times 2.51 \times 15 )
( 167200 - 3344 T_f = 1004 T_f - 15060 )
( 167200 + 15060 = 1004 T_f + 3344 T_f )
( 182260 = 4348 T_f )
( T_f = \frac{182260}{4348} \approx 41.9^\circ\text{C} )

Answer:

( \approx 41.9^\circ\text{C} ) (or ( 42.0^\circ\text{C} ) with rounding)