A 1000g sample of lead at 300 degrees C is dropped into 100 g of water at a temperature of 5.6 degrees C. The specific heats of lead and water are .129 and 4.184 J/g degC. What is the final temperature of the mixture?

Question

anonymous · Answer

$$(0.129)(1000)(T_{f} - 300°C) =-[ (4.184)(100)(T_{f} - 5.6)]\129T_{f} - 38700J = -[418.4T_{f}-2343.04J]\-38700J-2343.04J = -547.4T_{f}$$

anonymous · Answer

Solve

anonymous · Answer

1.23424333344 DEGREES

anonymous · Answer

ACCORDING TO MY CALCULATION THE MULTIPLICATION TO THE MOTHATRATIONS EQUAL THE CELIUS OF THE DEGREE ACCORDING TO PYTHAGOROUS

anonymous · Answer

Alright this kid above is on something else ignore him and solve...