Question

1. A piece of a metal (34.9 g) was heated to 95.0 degrees C and then added to 550-g water having an initial temperature of 28.2 degrees C. If the final temperature of the water and the metal is 31.2 degrees C, calculate the specific heat (s) of the metal. Assume not loss of energy to the surroundings! Show work!

2. A -10.0 mL solution of 0.320 M KOH was mixed with 25.0mL of 0.120 M HBr solution. Which one is the limiting reagent, KOH or HBr? Explain!

Answer #1

**1**.Sol :-

Heat gain by water (q_{1}) = Mass of water (w) x Heat
capacity of water (C) x Change in temperature (ΔT)

= 550 g x 4.184 J/g.^{0}C x (31.2 -
28.2)^{0}C

= 6903.6 J

Heat loss by metal (q_{2}) = Mass of metal (w) x Heat
capacity of metal (C) x Change in temperature (ΔT)

= 34.9 g x Heat capacity of metal (C) x (31.2 -
95.0)^{0}C

= 34.9 g x Heat capacity of metal (C) x (-
63.8)^{0}C

Negative sign indicates heat evolve.

Since, Heat gain by water (q_{1}) = Heat loss by metal
(q_{2})

So,

34.9 g x Heat capacity of metal (C) x ( 63.8)^{0}C =
6903.6 J

Heat capacity of metal (C) = 6903.6 J / (34.9 g x 63.8
^{0}C)

= 3.10 J/g.^{0}C

Hence, Heat capacity of metal = 3.10
J/g.^{0}C |

A -10.0 mL solution of 0.320 M KOH was mixed with 25.0mL of
0.120 M HBr solution. Which one is the limiting reagent, KOH or
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