Answer :
Let the final temperature be T.
Heat lost by water at 40°C is gained by water at 20°C. Thus
[tex]m_1s \Delta T_1=m_2s \Delta T_2\\ \\ 50 \times s \times (40-T)= 50 \times s \times (T-20)\\ \\40-T=T-20\\ \\T+T=40+20\\ \\2T=60\\ \\T= \frac{60}{2}=30^0C [/tex]
Final Temperature is 30°C
Heat lost by water at 40°C is gained by water at 20°C. Thus
[tex]m_1s \Delta T_1=m_2s \Delta T_2\\ \\ 50 \times s \times (40-T)= 50 \times s \times (T-20)\\ \\40-T=T-20\\ \\T+T=40+20\\ \\2T=60\\ \\T= \frac{60}{2}=30^0C [/tex]
Final Temperature is 30°C
by using the method of mixture formula,
final temperature T = M1×T1+M2×T2/ M1+M2
= 50×20 + 50×40/50+50
= 100+200/100
= 300/100
= 30 degrees c
therefore final temperature T = 30 degrees C
final temperature T = M1×T1+M2×T2/ M1+M2
= 50×20 + 50×40/50+50
= 100+200/100
= 300/100
= 30 degrees c
therefore final temperature T = 30 degrees C