Answer to Question #206109 in Mechanics | Relativity for marie

Question

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

Answer 1

$(1)$

Case first

$Q_1=1000J$

$T_1=300K$

$T_2=350K$

Mass =$m_1$

(2)Case second

$Q'_2=800J$

$T'_1=275K$

$T'_2=305K$

Mass =$m_2$

We know that

$Q=mc_v∆T$

Now

$Q_1=m_1c_v(T_2-T_1)\rightarrow(1)$

$Q'_2=m_2c_v(T'_2-T'_1)\rightarrow(2)$

equation (2) devided by equation (1)

$\frac{m_2}{m_1}=\frac{Q'_2}{Q_1}\times\frac{T_2-T_1}{T'_2-T'_1}$

Put value

$$\frac{m_2}{m_1}=\frac{800}{1000}\times\frac{350-300}{305-275}=\frac{8}{10}\times\frac{5}{3} =\frac{4}{3}$$

$\frac{m_2}{m_1}=\frac{4}{3}$