# 11. 135 g of aluminum (initially at "400"^("o")"C") are mixed with an unknown mass of water (initially at "25"^("o")"C"). When thermal equilibrium is reached, the system has a temperature of "80"^("o")"C". Find the mass of the water?

##### 1 Answer
Apr 7, 2015

about 170 g of water

The assumption that allows you to solve this problem is that the heat lost by the aluminum equals the heat gained by the water with no heat transfer outside the system

$- {Q}_{A l} = {Q}_{{H}_{2} O}$ where $Q = m c \Delta T$

-m_"Al"·c_"Al"·(T_f-T_i)=m_(H_2O)·c_(H_2O)(T_f-T_i)

substitution gives
$- \left(135 g\right) \left(0.9 J {g}^{- 1} {C}^{- 1}\right) \left({80}^{o} C - {400}^{o} C\right) = m \left(4.184 J {g}^{- 1} {C}^{- 1}\right) \left({80}^{o} C - {25}^{o} C\right)$
solve for m to get 169 g