Answer to Question #143589 in General Chemistry for Johannes Steven

Question

1. The freezing point of mercury is -38.8 degrees Celsius. What quantity of energy in joules is released to the surroundings of 1.00 mL of mercury is cooled from 23 degrees Celsius to -38.8 degrees Celsius and then frozen to a solid?

2. A 13.8 g of zinc is heated to 98.8 degrees Celsius and then dropped onto a beaker containing 45.0 g of water at 25.0 degrees Celsius. When the water and metal come to thermal equilibrium the temperature is 27.1 degrees Celsius. What is the specific heat capacity?

Answer 1

1. The density of liquid mercury is 13.6 g/cm³. Its specific heat capacity is 0.140 J/gK and its heat of fusion is 11.4 J/g.

$q_1 = mc_1∆T$ (due to temperature change)

c₁ – specific heat capacity

$m = ρ \times V$

ρ = 13.6 g/ml

$m = 13.6 \times 1 = 13.6 \;g$

c₁ = 0.140 J/gK

∆T = (23 - (-38.8)) = 61.8

$q_1 = 13.6 \times 0.140 \times 61.8 = 177.66 \;J$

$q_2 = mc_2$ (frozen to a solid)

c₂ – heat of fusion

c₂ = 11.4 J/g

$q_2 = 13.6 \times 11.4 = 155.04 \;J$

Total energy $q_T = q_1 + q_2$

$q_T = 177.66 + 155.04 = 332.7 \;J$

2. m(water) = 45 g

∆T(water) = 27.1 - 25 = 2.1 ºC

Specific heat capacity of water = 4.186 J

So, amount of heat absorbed by water q

q = mc∆T

$q = 45 \times 4.186 \times 2.1 = 395.577 \;J$

For zinc

m(Zn) = 13.8 g

∆T(Zn) = 100 - 27.1 = 72.9 ºC

Heat lost by zinc q

q = 395.577 J

q = mc∆T

Specific heat capacity of Zn

$c = \frac{q}{m∆T}$

$c = \frac{395.577}{13.8 \times 72.9} = 0.393 \;J/gC$