Question

The temperature of the water in a cup falls after the ice has been added. Specific heat capacity of water = 4200 J kg-1 K-1. (i) Calculate the mass of the water in the cup if the lowest temperature reached by the water is 8 degrees C.of the ice is

The energy to raise the temperature of the ice to 0 degrees C is 55(4) Joules

Answer 1

what is the start temperature of water?

Answer 2

The temperature of the water is 22 degrees C. To change the ice to water without an increase in temperature = 7260 J

Answer 3

ok just to sum it up:
start temperature 22°C
end temperature 8 °C
cp = 4,2 kJ kg-1K-1
DH of melting = 7,26 kJ

Answer 4

Yes. Also to to raise the temperature of the ice from -12 Degrees C to 0 Degrees C it takes 55(4) J

Answer 5

what is 55(4) i mean what is that (4)

Answer 6

Sorry. It should be 554.4 J

Answer 7

ok now it makes sense

Answer 8

just a sec to write it all

Answer 9

so you have ice at temperature -12 °C which is heated to 0 °C then you have water at 22°C which is cooled to 8°C and you have one phase transition ice to water
so lets start by calculating rise of temperature of ice
\[Q = m cp \Delta T\]
cp(ice) = 2,11 kJ kg-1 K-1
m= Q/cp DT = 554,4 J / 2,11 kJ K-1kg-1 * 12 K
m= 0,02189 kg = 21,89 g
now lets see what is the situation with liquid water:
m= Q/cpDT = 7260 J + 554,4J / 4200 J kg-1K-1 * 14K
m= 0,1328 kg = 132,8 g
and sumed up 154,69 g of water
now that is what i think your answer is

Answer 10

Convert the mass into Kg.

Answer 11

i think i made misstake

Answer 12

could you please write again complete problem in one peace cause i have to catch data all over the page...

Answer 13

Ok.This question is in two parts. Ice of mass 22g at temperature -12Degrees C is taken from a freezer and placed in a polystyrene cup containing water at a temperature 22 degrees C. (a) Calculate the quantity of heat needed (i) to raise the temperature of the ice from -12 degrees C to 0 degrees C (ii) to change the ice to water without an increase in temperature.

Answer 14

Then following on:The temperature of the water in a cup falls after the ice has been added. Specific heat capacity of water = 4200 J kg-1 K-1. (i) Calculate the mass of the water in the cup if the lowest temperature reached by the water is 8 degrees C.

Answer 15

ok and you need to be given heat capacity of ice and energy of phase transition

Answer 16

are you given heat capacity of ice and energy of phase transition?

Answer 17

Heat capacity of ice is 2100J kg-1 k-1. Specific latent heat of fusion of ice = 3.3 * 10^5 J kg-1.

Answer 18

ok now give me a minute to calculate it on paper and then ill write you all here

Answer 19

Ok

Answer 20

i got it

Answer 21

so first lets calculate heat that is needed for ice to be heated to 0°C
Q= mcpdT = 22*10-3 kg * 2100 J kg-1 K-1 * 12K = 554,4 J that is answer to (i)
then lets calculate heat that is needed to turn ice to water:
Q = m * DH = 22*10^-3 kg * 3,3 * 10^5 J kg-1 = 7260 J that is answer to (ii)
then lets calculate heat needed for water to be heated from 0 to 8 °C cause that sumed up heat is equal to heat of cooling water from 22 to 8 °C
Q= cp m dT = 4200 J kg-1 K-1 * 22*10^-3 kg * 8 K = 739,2 J
now lets sum up the heat:
Q = 739,2 + 7260 + 554,4 = 8553,6 J
and mass of water in cup is:
m= Q/ cp dT = 8553,6 J / 4200 J kg-1 K-1 * 14K = 0,1454 kg

Answer 22

Yes that is the correct answer!

Answer 23

im a bit rusty but my brain still works :)