Question #206109

A substance of mass m1 at a temperature of 300 K is supplied with 1000 J of energy such that its temperature increases to 350 K. When the same substance of mass m2 is supplied with 800J of energy, its temperature changes from 275 K to 305 K.


Determine the ratio of m1 to m2

Expert's answer

(1)(1)

Case first

Q1=1000JQ_1=1000J

T1=300KT_1=300K

T2=350KT_2=350K

Mass =m1m_1


(2)Case second

Q2′=800JQ'_2=800J

T1′=275KT'_1=275K

T2′=305KT'_2=305K

Mass =m2m_2

We know that

Q=mcv∆TQ=mc_v∆T

Now

Q1=m1cv(T2−T1)→(1)Q_1=m_1c_v(T_2-T_1)\rightarrow(1)

Q2′=m2cv(T2′−T1′)→(2)Q'_2=m_2c_v(T'_2-T'_1)\rightarrow(2)

equation (2) devided by equation (1)

m2m1=Q2′Q1×T2−T1T2′−T1′\frac{m_2}{m_1}=\frac{Q'_2}{Q_1}\times\frac{T_2-T_1}{T'_2-T'_1}

Put value


m2m1=8001000×350−300305−275=810×53=43\frac{m_2}{m_1}=\frac{800}{1000}\times\frac{350-300}{305-275}=\frac{8}{10}\times\frac{5}{3} =\frac{4}{3}

m2m1=43\frac{m_2}{m_1}=\frac{4}{3}


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