Answer to Question #218900 in Molecular Physics | Thermodynamics for EDUWA

Question

 the hot water tap of a bath delivers water at b5°c at a rate of 8kg/min. the cold water tap of the bath delivers water at 30°c at the rate of 20 kg/min. if both taps are left on for 5 minutes, calculate the final temperature of the bath water, ignoring heat losses to the surroundings

Answer 1

Gives

$T_1=5°c,\\\dot{m_1}=8kg/min \\\dot{m_2}=20kg/min \\t=5min \\T_2=30°c$

$\dot{Q_1}=\dot{m_1}cT_1$

Put value

$$\dot{Q_1}=8\times5\times 2108=84320J/min$$

$$\dot{Q_2}={20}\times30\times 4182=2509200J/min$$

Power loss

$= \dot{Q_2}-\dot{Q_1}=2424880J/min$

$∆\dot{Q}=2424.88kJ/min$

Heat loss

$∆{Q}=2424.88\times5=12124.4KJ$