Question

Equal mass of ice(-20тДГ) is mixed with water of temperature 80тДГ . What will be the temperature of the mixture ???

Answer 1

Answer:

Explanation:

•Like any problem in physics this can be solved in multiple manners

•For simplicity I’m going to consider only the thermodynamic branch of physics in this case

•The heat given and received by water and Ice respectively will be equal

•This idea will help us reach this equation

•MC1(T -(-20)) = MC2(80-T)

C1 ( T + 20 ) = C2 (80 - T)

TC1 + 20C1 = 80C2 - TC2

•T = 80C2 - 20C1 / C1 + C2

•Now simply plug in values of C1 and C2 :)

•Please mark as brainliest and feel free to ask for any clarification •