Ice
m₁=200 g = 0.2 kg
c₁ = 2060 J/kg•K
λ =335000 J/kg
Water
c₂=4183 J/kg•K
m₂=?
m₁c₁[0-(-5)] + m₁λ+ m₁c₂(21-0) = m₂c₂(65-21)
m₂=m₁(5c₁+λ+21c₂)/44 c₂ =
=0.2(5•2060 +335000+21•4183)/44•4183 =
=0.2(103000+335000+87843)/184052 =
=0.57 kg
200 g of ic is at -5C is dropped into a container containing water at 65C. If the final temperature of the mixture reaches 21C, what mass of water must be present in the container?
1 answer