Question

how do I do the calculations? please be very detailed, thank you?

Molar Heat of Fusion of Ice-Sample Worksheet Data: Mass of cup 5,50 g 155.50 g Mass of cup plus warm water 30.5 Initial temperature of "warm" water 0.0 C Initial temperature of ice 18.5 C Final temperature of the "mixture" 175.50 g Mass of cup plus "mixture" Results: Mass warm water Mass ice Temperature change of "warm" body Temperature change of "cool" body Heat energy lost by "warm" body Heat energy gained by "cool" body Heat of fusion per gram of ice Heat of fusion per mole of iceAD (Molar Heat of Fusion)

Answer 1

Date Fage No. Expt. No. Dat a Mass 5.508 Cup T J55.509 Mass c h twanm oahen Wader 20.5°C Intial tempenatue of anm 000c Tmhal tempenahun of the mixhune 18.5°c Pinal tempenahue 115.50 8 DMass of Cup + mixture AResults Mass of 1508 155.50-5.50 ef ice Mass 115.50-155 So= 2009 18 5-30.5 - 22.0 Temperchuee chamge of am bady Termpenatuae chang of 18.5-0-18 5°c Cold body Heal enngy last by H-mcdr bady Warm 150x 4.2x t220) X10 1o3 -1386 J Teoher's Signoture

Heat energy lost Joy ce 163 x 20x 4 2.1x 18.5 777 J Meat of fusion 1gm ofrce Lo 0.336 X106x 20x16 6.12X103 J Heat of fuion na L j'ce Pen mole 20X5 X 1434 6 JImele 18 1594 J 2
the value of specific heat capacity of water is 4.2kj/kg.K

Specific heat of ice = 2.1kj/kg.k

Latent heat of fusion of ice = 0.336MJ/kg

Molar latent heat of fusion of ice= 1434.6j/mole