Please show all steps. Constants Part A 1.00 kg of solid Hg at its melting point

Question

Please show all steps.

Constants Part A 1.00 kg of solid Hg at its melting point of-39°C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80 °C, the resulting equilibrium temperature is 5.06 °C. The specific heat of aluminium is 900 J/kg. C, specific heat of water is 4186 J/kg-C, specific heat of mercury is 138 J/kg.C Determine the latent heat of fusion of mercury using the given data Express your answer using three significant figures. Submit Request Answer Provide Feedback

Explanation / Answer

Given:
Mass of the solid mercury m1 = 1 kg
Mass of the aluminum container m2 = 0.620 kg
Mass of water m3 = 0.4 kg
Temperature of mercury is = - 39 oC
Temperature of aluminum + water t2 = 12.8 oC
Resultant temperature, t = 5.06 oC
Specific of mercury, c1 =138 J/kg.oC
Specific heat of aluminum, c2 = 900 J/kg.oC
Specific heat of water, c3 = 4186 J/kg.oC

Now, applying principle of calorimetry
m1 L + m1 c1 (t - t1 ) = m3 c3 ( t2 - t ) + m2 c2 ( t2 - t )
1 ( L) + (1 )( 138) ( 5.06 + 39 ) = (0.4) (4186 ) ( 12.8 - 5.06 ) + (0.620)( 900 ) ( 12.8 - 5.06 )
L + 6.08 x 10^3 = 1.3 x 10^4 + 4.32 x 10^3
L = 1.124 x 10^4 J/kg

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