Answer:
[tex]Q_a=330 cal[/tex]
[tex]Q_w=6000cal[/tex]
Explanation:
From the question we are told that
Mass of the aluminum container 50 g
Mass of the container and water 250 g
Mass of the water 200 g
Initial temperature of the container and water 20°C
Temperature of the steam 100°C
Final temperature of the container, water, and condensed steam 50°C
Mass of the container, water, and condensed steam 261 g
Mass of the steam 11 g Specific heat of aluminum 0.22 cal/g°C
a) Heat energy on container
Generally the formula for mathematically solving heat gain
[tex]Q_c=M_c *C_c*( \triangle T)[/tex]
Therefore imputing variables we have
[tex]Q_a=50g *0.22*50-20[/tex]
[tex]Q_a=330 cal[/tex]
b) Heat energy on water
Generally the formula for mathematically solving heat gain
[tex]Q_w=M_w *C_w*( \triangle T)[/tex]
Therefore imputing variables we have
[tex]Q_w=200 *1* 50-204[/tex]
[tex]Q_w=6000cal[/tex]