if 50g of ice were put into 200g of water at 75celsius what is the resulting temp?

Answer 1

put 50g of ice in 200g of water at 75celsius, now put a thermometer in the water, measure the resulting temp !!!!

its that simple

Answer 2

You have to take into account the heat required to melt the ice. This so-called heat of fusion is 80 calories/gram for water/ice. Melting the ice therefore takes 80 * 50 = 4000 calories. Since water has a heat capacity of 1 calories/(gram*degree C), this means the warm water must cool by 4000 / 200 = 20 degrees in order to provide the energy necessary for melting the ice.
Now we have 50 g of (liquid) water at 0 degrees, and 200 g at 75 - 20 = 55 degrees. From here we can just take a weighted average to find the final temperature: (0 * 50 + 55 * 200) / (50 + 200) = 44 degrees.

Answer 3

Assuming the ice is at 0degC, then you need to take into account the latent heat of fusion of ice (80cal per g) and get the equasion:

50*80 + 50(T-0) = 200(75-T); where T is the final temperature.

Solve for T -> 44degC.

Answer 4

60 degrees.

let final temperature be T.
heat gained = heat lost
50 * T= 200 (75-T)
50*T = 200*75 - 200*T
250 *T = 200 *75
T=60

Answer 5

if ice is at 0 celsius the resulting tem will be:
0.2* 4200*(75-resulting temp)+0.05*4200*(0-resulting temp)=0
33000-840 resulting temp-210 resulting temp=0
33000=1050 resulting temp
resulting temp=31.4
* multiply