Question

the specific heat of copper is 0.385 j/g°c. a 95.0 g block of copper is heated to 77.9°c and placed into a calorimeter that contains 42.0 ml of water at 20.4°c. assuming no heat loss, what will be the final temperature of the copper and water?
a 30.3°c
b 31.3°c
c 30.0°c
d 32.3°c

Explanation:

Step1: Determine the mass of water

The density of water is $1\ g/mL$. So, the mass of $42.0\ mL$ of water, $m_{w}=42.0\ g$.

Step2: Set up the heat - transfer equation

The heat lost by the copper is equal to the heat gained by the water. The heat - transfer formula is $Q = mc\Delta T$. Let the final temperature be $T$.
The heat lost by copper $Q_{cu}=m_{cu}c_{cu}(T_{cu}-T)$, and the heat gained by water $Q_{w}=m_{w}c_{w}(T - T_{w})$.
Since $Q_{cu}=Q_{w}$, we have $m_{cu}c_{cu}(T_{cu}-T)=m_{w}c_{w}(T - T_{w})$.
The specific heat of water $c_{w}=4.184\ J/g^{\circ}C$, $m_{cu}=95.0\ g$, $c_{cu}=0.385\ J/g^{\circ}C$, $T_{cu}=77.9^{\circ}C$, $m_{w}=42.0\ g$, and $T_{w}=20.4^{\circ}C$.
Substitute the values into the equation: $95.0\times0.385\times(77.9 - T)=42.0\times4.184\times(T - 20.4)$.

Step3: Expand the equation

$95.0\times0.385\times77.9-95.0\times0.385\times T=42.0\times4.184\times T-42.0\times4.184\times20.4$.
$95.0\times0.385\times77.9 = 95.0\times0.385\times T+42.0\times4.184\times T-42.0\times4.184\times20.4$.
$95.0\times0.385\times77.9+42.0\times4.184\times20.4=(95.0\times0.385 + 42.0\times4.184)T$.
First, calculate the left - hand side:
$95.0\times0.385\times77.9=95.0\times30.0715 = 2856.7925$.
$42.0\times4.184\times20.4 = 42.0\times85.3536=3584.8512$.
The sum of the left - hand side is $2856.7925+3584.8512 = 6441.6437$.
Then, calculate the coefficient of $T$ on the right - hand side:
$95.0\times0.385+42.0\times4.184=36.575+175.728 = 212.303$.

Step4: Solve for $T$

$T=\frac{6441.6437}{212.303}\approx30.3^{\circ}C$.

Answer:

A. $30.3^{\circ}C$