Question

a block of aluminum weighing 28.3 g is heated to 78.4 °c in boiling ethanol. it is then dropped into 33.8 g of ethanol initially at 12.5 °c. find the final temperature of aluminum + ethanol in °c. attempts: 0 of 6 used save for later etextbook and media submit answer current attempt in progress

Explanation:

Step1: Identify relevant specific - heat capacities

The specific - heat capacity of aluminum $c_{Al}=0.900\ J/g^{\circ}C$ and of ethanol $c_{ethanol}=2.46\ J/g^{\circ}C$.

Step2: Set up the heat - transfer equation

The heat lost by the aluminum is equal to the heat gained by the ethanol. The heat - transfer formula is $Q = mc\Delta T$. Let the final temperature be $T$.
The heat lost by aluminum $Q_{Al}=m_{Al}c_{Al}(T_{Al,i}-T)$, where $m_{Al} = 28.3\ g$, $c_{Al}=0.900\ J/g^{\circ}C$, and $T_{Al,i}=78.4^{\circ}C$.
The heat gained by ethanol $Q_{ethanol}=m_{ethanol}c_{ethanol}(T - T_{ethanol,i})$, where $m_{ethanol}=33.8\ g$, $c_{ethanol}=2.46\ J/g^{\circ}C$, and $T_{ethanol,i}=12.5^{\circ}C$.
Since $Q_{Al}=Q_{ethanol}$, we have $m_{Al}c_{Al}(T_{Al,i}-T)=m_{ethanol}c_{ethanol}(T - T_{ethanol,i})$.

Step3: Expand the equation

$28.3\times0.900\times(78.4 - T)=33.8\times2.46\times(T - 12.5)$.
First, calculate $28.3\times0.900 = 25.47$ and $33.8\times2.46=83.148$.
The equation becomes $25.47\times(78.4 - T)=83.148\times(T - 12.5)$.
Expand both sides: $25.47\times78.4-25.47T=83.148T-83.148\times12.5$.
$25.47\times78.4 = 1996.848$ and $83.148\times12.5 = 1039.35$.
So, $1996.848-25.47T=83.148T - 1039.35$.

Step4: Solve for $T$

Add $25.47T$ to both sides: $1996.848=83.148T + 25.47T-1039.35$.
$1996.848 + 1039.35=(83.148 + 25.47)T$.
$3036.198 = 108.618T$.
$T=\frac{3036.198}{108.618}\approx27.95^{\circ}C$.

Answer:

$27.95^{\circ}C$