Answer to Question #156877 in General Chemistry for H

Question

1.     105.0 mL of H₂O is initially at room temperature (22.0 °C). A chilled steel rod (2.00 °C) is placed in the water. If the final temperature of the system is 21.5 °C, what is the mass of the steel bar? (specific heat of water = 4.184 J/g °C; specific heat of steel = 0.452 J/g °C).

Answer 1

Volume of water, V = 105 mL

Mass of $H_2O$, m = 105 * 1.00 g= 105g

Initial temperature, $T_i$ = 22.0 °C

Final temperature, $T_f$ = 21.5 °C

Specific heat of water = 4.184 J/g °C

Heat energy, q = 105 g* 4.184 J/g °C (21.5 °C - 22.0 °C) = - 219.66 J

Let mass of the steel rod is x

Initial temperature, $T_i$ = 2.0 °C

Final temperature, $T_f$ = 21.5 °C

Specific heat of steel = 0.452 J/g °C

Heat energy, q = x * 0.452 J/g °C (21.5 °C - 2.0 °C) = 8.814x

But, heat energy lost by rod is equal to heat energy gained by water. So,

$-q_r(lost) = q_w(gain)$

-8.814x = - 219.66 J

x = 219.66/8.814 = 24.92

Hence, the mass of steel bar is 24.92g