The figure below is a heat curve for an unknown outerspace substance discovered

Question

The figure below is a heat curve for an unknown outerspace substance discovered by the Mars rover By understanding the heat curve, scientists can make inferences about the composition of the planet. , and vaporization, Tvapor, for this unknown substance? What is the temperature of fusion, Tfusion Note that the units are graded Temperature (C) 400 Number Units Tfusion = ID 320 Units 180 160 140 120 vapor For a sample of 88.2 g of this substance with a specific heat of 46.4 J/kg. °C, how much heat is added to the substance if the temperature is to change from 180.0°C to 256.0? 40 80 120 Number Time

Explanation / Answer

a)

Tfussion --> must be the sable temperature at which solid goes to liquid, from the graph, it is approx

T fus = 80°C

b)

Tvapor --> it is the temperature require to change form liquid to vapor, it must be higher than T fusion

this is T = 280°C approx

c)

m = 88.2 g of substance

Cp = 46.4 J/kgC

find het for:

180°C to 256°C

This is a liquid

therefore

qliq = mliq*Cpliq(Tf-Ti)

Qliq = 88.2*46.4*(256-180) = 311028.48 J

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