# Calculate the heat energy released when 20.1 g of liquid mercury at 25.00 °C is converted to (s) mercury at its melting point? Constants for mercury at 1 atm: heat capacity of Hg(l)=28.0 J(mol*K), melting point=234.32 K, enthalpy of fusion=2.29 KJ/mol

You’d release 409 J when converting that much mercury from liquid to solid at its melting point.

There are two steps to bringing the mercury from liquid to solid at melting point – you must first get the mercury from liquid at room temeprature to liquid at melting point, and then from liquid at melting point to solid at melting point.

Notice that you were given the molar heat capacity and the molar heat of fusion of mercury, so it’ll be easier to work with moles instead of grams. The number of moles of mercury can be determined using its molar mass

##”20.1 g” * “1 mole”https://studydaddy.com/”200.59 g” = “0.1002 moles Hg”##

The first step will get you from liquid mercury at 25.00 + 273.15 = 298.15 K to liquid mercury at 234.22 K.

##q_1 = n * c_(“molar”) * DeltaT##

##q_1 = “0.1002 moles” * 28.0″J”/(“mol” * “K”) * (234.22 – 298.15)”K”##

##q_1 = “-179.08 J”##

Now for the second part, which represents one of mercury’s . Notice that you were given the heat of fusion, which is the heat required to change one gram of a substance from solid to liquid. Since you’re going the opposite way, from liquid to solid, the heat of fusion will have a negative sign because heat is released, not put into this process.

This time, the energy needed will be

##q_2 = n * DeltaH_(“f molar”)##

##q_2 = “0.1002 moles” * (-2.29″kJ”https://studydaddy.com/”mol”) = “-0.22946 kJ” = “-229.46 J”##

The total heat released for this particular process will be

##q_(“TOTAL”) = q_1 + q_2 = “-179.08 J” + (“-229.46 J”) = “-408.54 J”##

Rounded to three , the same number of sig figs in 20.1 g, the answer will be

##q_(“TOTAL”) = “-409 J”##

SIDE NOTE You can try and solve this problem by converting the molar heat capacity to capacity (to kJ per g K), and the molar heat of fusion to the heat of fusion per gram (kJ/g).

The result should be the same.

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