Pls help me God will bless you all. A piece of copper ball of mass 20 g at 200 degree Celsius is placed in a 50 g of water at 30 degree Celsius, ignoring heat losses, calculate the final steady temperature of the mixture. Specific heat capacity of water=4.2j/ kg.g

Answer 1

sum of heats gained = zero.
.020*specheatCu*(Tf-200) +50*4.2*(Tf-30)=0
solve for Tf. Watch units