Answer
a) $9.12\cdot10^5\ J$
b) $11.8°C$
Work Step by Step
a) Use the sensible heat equation for the rocks to calculate the heat:
$q=mc\Delta T$
$q=(69.7\ kg\cdot\frac{1\ kg}{10^3\ g})\cdot 0.818\ J/g.°C\cdot (41.0°C-25.0°C)$
$q=9.12\cdot10^5\ J$
b) First, calculate the mass of air in g:
$d=\frac mV$
$m=(2.83\cdot10^5\ L\cdot \frac{10^3\ mL}{1\ L})\cdot 1.20\cdot10^{-3}\ g/mL=3.40\cdot10^5\ g$
Calculate the heat released by the rocks while cooling from 41.0°C to 30°C:
$q=mc\Delta T$
$q_r=(69.7\ kg\cdot\frac{1\ kg}{10^3\ g})\cdot 0.818\ J/g.°C\cdot (30.0°C-41.0°C)$
$q_r=-6.27\cdot10^5\ J$
Neglecting heat losses, all the heat lost by the rocks is absorbed by the air:
$q_r+q_a=0$
$q_a=+6.27\cdot10^5\ J$
Calculate the final temperature of the air from the sensible heat equation:
$q=mc\Delta T$
$6.27\cdot10^5\ J=3.40\cdot10^5\ g\cdot 1.004\ J/g.°C\cdot (T_f-10°C)$
$1.84=T_f-10$
$T_f=11.8°C$
