what mass of ice at -14'c will be needed to cool 200cm3 of an orange drink (essentially water) from 25'c to 10'c! (specific latent heat of fusion of ice =336000) (specific heat capacity of ice = 2100) (specific heat

Answer 1

No units given. I looked them up.
latent heat fusion ice = 334 J/g*C
specific heat ice = 2.1 J/g*C
Step 1. heat ice from -14 to zero.
Step 2. melt ice @ zero to liquid water @ zero
Step 3. heat melted ice from zero to 10 C.
All of that must equal the 200 cc orange drink going from 25 C to 10 C (delta T = 15 )
Remember q = mcdT
step 1. m*2.1*14
step 2. m*heat fusion = m*334
step 3. m*4.18*10.
orange drink loses mcdT = 200*4.18*15
heat added to ice = heat lost from orange drink
(m*2.1*14) + (m*334) + (m*4.18*10) = 200*4.18*15
Solve for m = mass ice.
I estimate between 30 and 35 g ice @ -14 C.
Post your work if you get stuck.

Answer 2

I used a lot of different values for the parameters. Like the SLH and SHC. I got 36kg but I'm kinda confused cuz they said the answer is something like 0.033kg