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. What is the temperature of fusion, Tusion. and vaporization, Tvpor, for this unknown substance? Note that the units are graded. Temperature (C) 400 330 360 340 320 Number Units = fusion 280 260 240 220 Number Units 130 160 apur 120 30 20 40 30 For a sample of 54.6 g of this substance with a specific heat of 52.0 J/kg C, how much heat is added to the substance if the temperature is to change from 180.0 C to 220.0°C? 30 Time 120 Number

Explanation / Answer

m = 54.6 g

Cp = 52 J/oCg

T1 = 180 oC

T2 = 220 oC

Q = m Cp dT

Q = 54.6 x 52 x (220-180)

Q =113568 J

2)

100 g water can dissolve ----------------------> 36.9 g   at 20oC

60 g water can dissolve --------------------> 60 x 36.9 /100 = 22.14 g at 20oC

100g can dissolve -------------------> 84.5 g at 30oC

60g water ----------------------------->   60 x 84.5 / 100 =50.7 g at 30oC

solute dissolved = 50.7 -22.14 = 28.56 g

28.56 more solute can add at 30oC

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