Algebra help please?

Question

A chemist mixes 300 grams of water at 75 degrees Celsius with 100 grams of water at 15 degrees Celsius. To find the final temperature of the water after mixing, use the equation m1(t1-t)=m2(t-t2), where m1 is the quantity of water at the hotter temperature, t1 is the temperature of the hotter water, m2 is the quantity of water at the cooler temperature, t2 is the cooler temperature of the cooler water, and t is the final temperature of water after mixing.

Answer 1

let the temperature of the mixture be t
300(75-t)=100(t-15)
22500-300t=100t-1500
adding 1500
24000-300t=100t
adding 300t
24000=400t
dividing by 400
t=60

Answer 2

Using the equation you've given:

m1(t1 - t) = m2(t - t2)
300(75 - t) = 100(t - 15)
22,500 - 300t = 100t - 1500
400t= 24,000
t = 60°C

Answer 3

i really cant understand what you want.. but i guess you want see if this equation works? or may be the final temperature,, who knows? anyway if you mixed 2 quantities of cold and hot water that can make the mixture warm but temperature will be another case as water gain and lose heat fast, so if you done that in isothermal thormos may be you wont lost that much of temperature but ofcourse it wont be the theortical temperature you will calculate with your equation

Answer 4

The problem gave you the formula, and all the values you need. Substitute in the values into the formula, then solve for t. You should be able to do this.