Question

if 80 gram of water at 45 degree celcius temperature is added to 20 gram of water at 30 degree celcius temprature. Then what will be the temprature of mixture​

Answer 1

Answer :-

$\bf{m _1 = 80 g, T_1 = 45 °C, m_2 = 20 g,} \\ \sf T_2 = 30 °C, T = ?$

According to the principle of heat exchange, heat lost by hot water = heat gained by cold water

$\sf∴ m_1 \: c (T1 – T) = m_2 \: c (T – T_2)$

$\sf∴ m_1T_1 – m_1T = m_2T – m_2T_2$

$\sf∴ m_1T_1 + m_2T_2 = (m_1 + m_2)T$

∴ Maximum temperature of the mixture

$\sf \: T = m_1T_1 + m_2T_2 m_1+m_2$

$= \large \sf \frac{80g + 45°C + 20g × 30°C }{80g + 20g}$

$= \large \sf( \frac{80×45}{100} + \frac{20×30}{100} )$

= (36 + 6) °C

= 42°C

$\large\boxed{\fcolorbox{pink}{ink}{Hope it helps uh :)}}$

Answer 2

Answer:

100 gram water at 75C temperature.