Physics, asked by morankhiraj, 3 months ago


 \footnotesize\sf \red{A \:  body  \: of  \: mass  \: "m_{1}"  \: of  \: a  \: substance \: of \: specfic \: head} \\   \footnotesize\sf \red{capacity \:  \:  \: c_1, \: at \: a \: \:  \:  temperature \: t_1 \:  \:  \: i s \:  \: mixed \:  \:  \:  \: with} \\  \footnotesize \sf \red{another \: body \: of \: mass \:  "m_2"\: of \: specific \: head \: capacity} \\  \footnotesize \sf \red{"c_2" \: at \: a \: lower \: temperature \: t_2 . \: Deduce \:  an  \: expression} \\   \footnotesize \sf \red{for \:  \:  \:  \:  the \:  \:  \:  \:  temperature  \:  \:  \:  \:  \:  \: of  \:  \:  \:  \:  \: \:  the \:  \:  \:  \:  \:  mixture  \:  \: \:  t_3. }

Answers

Answered by Anonymous
1

Answer:

Heat lost = Heat gain

» m1 c1 (t1 - t3) = m2 c2 (t3 - t2)

» t1 (m1 c1) - t3 (m1 c1) =

t3 (m2 c2) - t2 (m2 c2)

» t1 (m1 c1) + t2 (m2 c2) =

t3 (m2 c2) + t3 (m1 c1)

» t3

= [t1 (m1 c1) + t2 (m2 c2)] / (m1 c1 + m2 c2)

Is the required expression.

Answered by gurjarvratika18
1

Answer:

ki vajah se face change ho gaya

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