Answer to Question #77757 in Molecular Physics | Thermodynamics for Martha Koroma

Question

Some hot water was added to 3times it's mass of water at 10oC and the result in temperature was 20oC. What was the temperature of the hot water?

Answer 1

As we know, at the end of experiment temperatures of cold and hot water will be equal. Cold water will get Q_1 j. and hot water will give Q_2 j. Thermodynamics laws says, that Q_1=Q_2.
So: m_1 c(t_0-t_1 )=m_2 c(t_2-t_0 ), where m_1 - is a mass of cold water, m_2 – is a mass of hot water, c – is specific heat of water, t_1- is temperature of cold water, t_2- is temperature of hot water t_0- is temperature of mixture.
m_1 c(t_0-t_1 )=m_2 c(t_2-t_0 )
〖3m〗_2 (t_0-t_1 )=m_2 (t_2-t_0 )
3t_0-〖3t〗_1=t_2-t_0
t_0+3t_0-3t_1=t_2
t_2=〖4t〗_0-3t_1
t_0=20C, t_1=10C, so
t_2=4×20-3×10=80-30=50℃
Temperature of the hot water is t_2=50℃.