Question

A 500.0-g sample of an element at 153°C is dropped into an ice-water mixture; 109.5-g of ice melts and an ice-water mixture remains. Calculate the specific heat of the element from the following data:

Specific heat capacity of ice: 2.03 J/g-°C

Specific heat capacity of water: 4.18 J/g-°C

H2O (s) → H2O (l), ΔHfusion: 6.02 kJ/mol (at 0°C)

a) If the molar heat capacity of the metal is 26.31 J/mol-°C, what is the molar mass of the metal, and what do you think it is?

Answer #1

q = nL fustion

n = 109.5/18 = 6.083moles

q = 6.083*6.02 = 36.619KJ = 36619J

for metal

q = mCDT

q = 500*C*(153-0)

Heat lose of metal = heat gain of water

36619 = 500*C*153

C = 36619/500*153 = 0.478J/g-c^0

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