Can someone help me figure out question b? A 0.300kg sample is placed in a cooli

Question

Can someone help me figure out question b?

A 0.300kg sample is placed in a cooling apparatus that removes energy heat a constant rate. The figure below during the temperature T of the sample versus time T; the horizontal scale by ts=88.0min. The sample freezes during removal . The specific heat of the sample in its initial liquid phase is 2640j/kg. k What is the sample's heat of fusion? Do you that(1) the temperature of the liquid decreases,(2) then the material freezes ?For the first stage , did you take a derivates of both sides relating energy removal temperature change? How is the latter related to the steps of the graph in the stage of cooling? The same rate of energy removal applies to stages 2 and 3.

Explanation / Answer

Heat taken away in cooling phase=q*44 where q is the constant heat removal rate.

44q=0.3*2640*(30)

q=540J/min

Heat lost in phase 3=q*ts/4=22q=11880J

0.3*Cp*20=11880

Cp=1980J/kg.K

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