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℃.