Demonstrating the low melting point of Gallium metal (29.8C), an instructor remo

Question

Demonstrating the low melting point of Gallium metal (29.8C), an instructor removes a 1.00 kg sample of the metal from a container at room temperature (20.0C) and places it onto the palm of his hand. Given the heat of fusion and specific heat of this metal is 80.18 kJ/kg and 373 J/kgC, respectively, determine the amount of thermal energy absorbed by the sample before reaching thermal equilibrium with the instructors hand (37.0C). Assume the specific heat of gallium does not change from the solid state to the liquid state.

Explanation / Answer

on reaching from 20 to 29.8 C : heat absorbed = m.C.dT = 1 x 373 x (29.8 - 20) = Q1 = 3655.4 KJ

during melring heat absorbed = Q2 = m.h = 1 x 80.18 = 80.18 KJ

on reaching from 29.8 to 37 , heat absorbed = Q3 = m.C.dT = 1 x 373 x (37 - 29.8) = 2685.6 KJ

so total thermal energy absorbed = Q1 + Q2 + Q3 = 3655.4 + 80.18 + 2685.6 = 26318.88 KJ

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