Question

What is the amount of heat (in Joules) absorbed by $18 \, g$ of water initially at room temperature heated to $100^{\circ} C$ ? If $10\, g$ of $Cu$ is added to this water, than decrease in temperature (in Kelvin) of water was found to be? $C(p, m)$ for water $75.32 \, J / mol K ; C(p, m)$ for $Cu =24.47 \, J / mol K$.

• A. 5649,369
• B. 5544,324
• C. 5278,342
• D. 3425,425

Answer 1

Answer: (A)

$18 \, g$ of water at $100^{\circ} C$
$10 \, g$ of $Cu$ at $25^{\circ} C$ is added.
$q_p=C_{p, m} dT$

$=75.32 \times \frac{ J }{ K mol } \times \frac{18\, g }{18\, g / mol }(373-298) K$
$=75.32 \frac{ J }{ K } \times 75 K =5649\, J$
If now $10 \, g$ of copper is added $C_{p, m }=24.47 \, J / mol K$
Amount of heat gained by $Cu$
$=24.47 \frac{ J }{ K mol } \times \frac{10\, g }{63\, g / mol }(373-298) K =291.3 J$
Heat lost by water $=291.30 J$
$-291.30\, J =75.32 \frac{ J }{ K } \times\left(T_2-373 K \right) $
$\Rightarrow -3.947 \, K =T_2-373 K$
$\Rightarrow T_2=369.05 \,K$