ElectricPavlov

Just multiply it out    379 x  . 28         Use the calculator.....

(BTW : I do not see the correct answer....did you enter answer "B" correctly? )

(x+10)(x-6) = 0

Of note.....the water will have to be under pressure or some of it will turn to steam.

If steam is allowed then the max temperature reached will be 100 degrees C and some of the water will be turned to steam at 100 degrees

How much?

The latent heat of vaoprization of water is 2230 j/gm   so if we are making steam the equation becomes:

300(.38)(800- 100) = 50 (2.05)(25) + 50(334) + 50(4.188)(100)  +  Wm(2230)   where Wm = mass of water turned to steam

Wm = 17.75 gms of steam and the entire system remains at 100 C

Assuming 25 degrees (room temperature)  Kw = 14  (self ionization of water)

[OH-]  =  Kw/[H+]  =  14/(2 x 10^-3) =  7000      I think!

Something is wrong here.....  divide both sides by n and you get   :    .10 = 50      No Go !

Maybe you mean    .1 n = 50     then n= 50/.1 = 500

-43 degrees     or    317 degrees   or   223 degrees

Looks like we are just adding 3/4 sequentially       n (3/4)

3 , 3 3/4,  41/2 , 5 1/4 .....etc

Again, the heat the brass ball LOSES is GAINED  by the ICE (s=2.05)and the WATER (s=4.188) that forms from the melting ice  AND some of the heat is needed to MELT the ice (Latent heat of fusion (or melting) of ice is 334 j/gm)

Brass heat loss =    300 (.38) (800- Tf)      where Tf is the final temp

Heat GAINED by ice      50 (2.05)(25)        25 is the temp change to get the ice to 0 degrees where it melts

         Melting ice             50(334)

Heat gained by the WATER (after the ice melts) =  50(4.188)(Tf)        (since the water warms from 0 degrees to Tf

now equate the losses and gains

300(.38)(800- Tf) = 50 (2.05)(25) + 50(334) + 50(4.188)(Tf)

Solve for Tf

91,200 - 114Tf = 2,562.5 +16,700 + 209.4Tf

71937.5 = 323.4 Tf

Tf = 222.44 degrees C

I depends on the VOLTAGE and number of turns in the PRIMARY coil ....

Yes, I do.

The heat the copper loses is GAINED by the water

specific heat of water = 4.186 j/(gm deltaT)

Let m= mass of copper

m (.385) (400-58) = 30(4.186)(58-18)   solve for m

m=  30(4.186)(40) / ((.385)(342))

m= 38.15 gm