ElectricPavlov

Thanx GA...I have always used 80  

Answer already posted to previous submission.

I apologize.....I used the wrong specific heat for ice   I use 1     it is .5

so to heat the ice to zero degrees will require

300 x .5 x 15 = 2250 cal    (not 4500)

NOW add them up   2250 + 24000 + 6000 = 32250 cal = 134934 j = 134.9 kj

Sorry!

Maybe a little easier way

16 ^2t  = n ^4t

(4^2)^2t = n^4t

4^4t = n^4t         so n = 4

Miriam starts with   16^2t      And J starts with n^4t    and we want these to be equal.....

Sooooo let's set them equal and solve for n    the amount of bacteria Jessica needs

16^2t  = n ^4t         Log both sides to get

2t log16  = 4t log n  divide by 2t

log16 = 2 log n        divide by 2

(log 16 )/2 = log n    10 power both sides

10^((log16)/2) = n = 4   the number J needs to start with to keep things equal

....and the second part....y'all must be using the same home schooling programs!

https://web2.0calc.com/questions/part-of-a-two-column-proof-is-shown-with-just-the_1

Here is the first part from yesterday:

https://web2.0calc.com/questions/a-right-triangle-is-shown-below

First let's find out how many CALORIES are required, and then convert to joules , knowing

1 cal = 4.184 j

One calorie will raise ONE GRAM of water ONE DEGREE Celsius

#calories = 7000 gm  x 78 C = 546000  CALORIES

546000 calories x 4.184 j/calorie = 2284.5 kj    (kj = KILOjoule)

2.  First you have to warm the ice up to 0 degrees

     300 g x 1cal/gram-C x  15 C = 4500 cal

     Then you have to add heat to it AT A CONSTANT TEMP to melt the ice @ 80 cal/gm

300gm x 80 cal/gm = 24000 cal

     Then you have to warm the water (melted ice at 0 degrees C) to 20 C

300 gm x 1 cal/gram-C  x 20 C  = 6000 calories

Add them all up  4500+24000+6000 = 34500 cal     =    144.3 kj

Thanx for the correction GA !     ~EP

(x-5)   *   (x+5)       are the two numbers multiplied together,,......