A 46 g ice cube at 0◦C is placed in 988 g of water at 90◦C. What is the final temperature of the mixture? The specific heat of water is 4186 J/kg · ◦C and its latent heat of fusion is 3.33 × 105 J/kg .

Question

A 46 g ice cube at 0◦C is placed in 988 g of water at 90◦C. What is the final temperature of the mixture? The specific heat of water is 4186 J/kg · ◦C and its latent heat of fusion is 3.33 × 105 J/kg .
Answer in units of ◦C

Elena · Answer

m(ice)=m₁=0.046 kg L=3.33•10⁵J/kg m(water) =m₂=0.988 kg c=4186 J/kg•℃ m₁L+m₁c(t-0) =m₂c(90-t) Solve for t