I believe you forgot to include some important information for this question.
How many grams of ICE would it take to drop the temperature of 150-g of WATER from 24.3 to 23.6? (using a polystyrene coffee cup calorimeter at room temperature)
How can I calculate the H-fusion of ICE from the above data?
How many grams of ICE would it take to drop the temperature of 150-g of WATER from 24.3 to 23.6? (using a polystyrene coffee cup calorimeter at room temperature)
How can I calculate the H-fusion of ICE from the above data?
Lets do this problem experimentaly😀
DQ = m c DT
D= Delta
The specific heat of water is 1 cal / g K by definition. That of ice is .51
Water first
DT= 0.7
c is given
m = 150g
so Q for needed for water is 105
now for the ICE
DT= im guessing its 23.6 since ice is 0 at initial and you want to goto 23.6
c is given
m is what you need
Q =105
so do plug in the info and solve for m
i get 8.72g
did i do it right?
How many grams of ICE would it take to drop the temperature of 150-g of WATER from 24.3 to 23.6? (using a polystyrene coffee cup calorimeter at room temperature)
How can I calculate the H-fusion of ICE from the above data?
The heat of fusion for water (ice) is 80cal/g.