Question

540g of ice at 0 celcius is mixed with 540g of water at 80ce

Answer 1

Answer:

tf=0.086°C

Explanation:

mL+mcΔt=−mcΔt

(540g)(334Jg)+(540g)⋅(4.184Jg°C)(tf−0°)=−(540g)(4.184Jg°C)(tf−80°)

180360J+2259.36J°C⋅tf−0J=−2259.36J°C+180748.8J

4518.72J°C⋅tf=388.8J

tf=0.086°C