plzzzzzzzzz help 4 chemistry/////i beg u..?

Question

Q1)) 3kg of ice is dropped in 9kg of water at 55 degree C. Find the final temprature of the mixture.???

Answer 1

I will assume that temperature of ice is 0°C and room temperature is 20°C

m1C1T1 + m2C2T2 = mCT

(3Kg)(1 kcal/kg°C)(20-0) + (9 kg)(1 kcal/kg°C)(55°-20°C) = (12 kg)(1 kcal /kg°C)(T)
60 + 315 = 12 T
T = 31.25°C

Hope it helps! (At last the methodology)

Good luck!

Answer 2

You will need to know the heat capacity of both water and ice.

Use the equation
q = n x Cn x (Tf - Ti)
where
q = amount of heat energy gained or lost by substance
n = moles of substance
Cn = molar heat capacity (J oC-1 mol-1 or J K-1 mol-1)
Tf = final temperature
Ti = initial temperature

for this question, you will need the Cn for both ice and water as well.
For ice, Cnice = 38.09 J K-1 mol-1
and water, CnH2O = 75.327 J K-1 mol-1

since the heat lost by water must be equal to the heat gain by ice, therefore
nH2O x CnH2O x (55 - Tf) = nice x Cnice x (Tf - 0)
(I'm assuming the initial temperature of ice at 0oC)
nH2O = 3/18
nice = 9/18
Sub it back into the above equation to find Tf.
Pardon me for being too lazy to press my calculator.

Answer 3

The Temperature will decrease therefore the temp will be less the 55 degree C.....

Answer 4

What is the ambient temperature? Is it one block of ice or cubes or what?

Answer 5

What is the temp. of the ice ?????????