Respuesta :
Problem: What is the most likely final temperature of the mixture
Given: m1 =400ml m2=100ml
c1= 4.18J/g∘C (specific heat of water) c2=4.18J/g∘C
T1=30C T2=80C
Solution: Formula: T = ( m1c1T1 + m2c2T2 ) / ( m1c1 + m2c2 )
[(400)(4.18)(30)+(100)(4,18)(80)] /[(400)(4.18)+(100)(4.18)]
=[(50,160)+(33,440)] / [(1672)+(418)]
=83,600/2090
Answer =40C
Given: m1 =400ml m2=100ml
c1= 4.18J/g∘C (specific heat of water) c2=4.18J/g∘C
T1=30C T2=80C
Solution: Formula: T = ( m1c1T1 + m2c2T2 ) / ( m1c1 + m2c2 )
[(400)(4.18)(30)+(100)(4,18)(80)] /[(400)(4.18)+(100)(4.18)]
=[(50,160)+(33,440)] / [(1672)+(418)]
=83,600/2090
Answer =40C