A 75g ice cube at 0 degree celcius is placed in 825 g of water at 25 degree celcius. What is the final temperature of the mixture.
specific heat of water = 4186 j/kg
Latent heat of water 3.33 x 10^5
formula
Q= L x m
please show step by step.
How can I find the temperature if there is no T in the formula? am I supposed to find the heat(Q) released from 0 degree celcius ice to 0 degree celcius water first using Q=Lm and then from 0 degree water to 25 degree water with Q= m C delta T to find T?
I did that but did NOT get the right answer
right answer is 16.3 degree celcius
specific heat of water = 4186 j/kg
Latent heat of water 3.33 x 10^5
formula
Q= L x m
please show step by step.
How can I find the temperature if there is no T in the formula? am I supposed to find the heat(Q) released from 0 degree celcius ice to 0 degree celcius water first using Q=Lm and then from 0 degree water to 25 degree water with Q= m C delta T to find T?
I did that but did NOT get the right answer
right answer is 16.3 degree celcius
-
You are using the correct formulas. Check to make sure your units are correct. Are the units for latent heat and specific heat correct?
The amount of heat needed to raise 75 grams of ice to X degrees is (75grams * latent heat) + (75grams * specific heat of water * delta T) where delta T = X- 0.0
This amount of heat is equal to the amount of heat removed from the 825 gram water bath to bring the temperature to X degrees. delta T = 25- X, m = 825grams
The amount of heat needed to raise 75 grams of ice to X degrees is (75grams * latent heat) + (75grams * specific heat of water * delta T) where delta T = X- 0.0
This amount of heat is equal to the amount of heat removed from the 825 gram water bath to bring the temperature to X degrees. delta T = 25- X, m = 825grams