Question

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

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

Answer 1

Answer: (A)

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

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