Question

We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s 50 C. We decided that it would be more convenient to drop an ice (0 C) since it will get cold soon. What mass of ice needs to be put in for this to happen?

Answer 1

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0