Jose mixes 1 gram of water at 0 celsius with 10 grams of water at 100 celsius. What is the final temperature?

Answer 1

Since it's all liquid water (no ice or vapor), you can just do a weighted average. [(1 g)*(0 C) + (10 g)(100 C)] / (11 g) = 90.91 C.

If you're supposed to be more formal about it, you need to set the final temperature as a variable, like T, and then calculate the heat gained by the 1 g at 0 C to be equal to the heat lost by the 10 g at 100 C. It would look like (1 g)c(T - 0) = (10 g)c(100 - T), where c is the specific heat capacity of water. As you can see, c cancels out from both sides, giving you T = 1000 - 10T ==> 11T = 1000 ==> T = 1000/11 = 90.91 C, as described above.

Answer 2

The specific heat of water is 1cal/g Lets tf the final temperatue
the hot water loses an amount of calories x

Q1 =- 10 (100-tf) =-1000+10tf

the cold water receives Q2 = 1*tf

Q1 =-Q2

-1000 +10tf = -tf

1000 = 11tf tf = 1000/11=90.9°C

Answer 3

(1*0+10*100)/11=90.9 celsius

Answer 4

the particular warmth of water is 1cal/g/°C. (or 1kcal/kg/°C). meaning, that a million gram of water + a million Calorie = a million°C improve in temperature. (get rid of a million calorie and the temperature is going down by a million°C) 1kilo-gram (a million,000 grams) demands 1kilo-calorie (a million,000 energy) in line with °C. those are 'small energy' 'c'. 'C' is 1kcal utilized by dieticians.

Answer 5

Jose needs to go back to mexico.

Answer 6

90.9 celsius

Answer 7

simple answer is 90..........i aint doing your homework for you and showing you how thou