Question

Hey,
I have been trying to figure out what I need to do for this lab report for about an hour and 1/2 and can't figure out what I'm supposed to do. I have looked through my notes and the lessons thoroughly. Please help!!! Here is one of the questions I need help with.1. For each of your three trials state the following:
• heat needed to melt the ice (q)
• enthalpy of fusion
• percent error from the accepted enthalpy of fusion of water of 334 J/g
I will post my data in a separate post!

Answer 1

Here is my Data!
Trial 1
Mass of calorimeter: 409g
Temp of calorimeter: 20Cº
Mass of calorimeter + water: 457g
Temp of calorimeter + water: 20Cº
Mass of calorimeter + water + ice: 510g
Temp of calorimeter + water + ice: –
Mass of calorimeter + ice: 450g
Temp of calorimeter + ice: 0Cº

Trial 2
Mass of calorimeter: 409g
Temp of calorimeter: 20Cº
Mass of calorimeter + water: 461g
Temp of calorimeter + water: 22Cº
Mass of calorimeter + water + ice: 515g
Temp of calorimeter + water + ice: –
Mass of calorimeter + ice: 447g
Temp of calorimeter + ice: 0Cº

Trial 3
Mass of calorimeter: 409g
Temp of calorimeter: 20Cº
Mass of calorimeter + water: 459g
Temp of calorimeter + water: 22Cº
Mass of calorimeter + water + ice: 513g
Temp of calorimeter + water + ice: –
Mass of calorimeter + ice: 448g
Temp of calorimeter + ice: 0Cº

Answer 2

you have to use the calorimetry formula: \(\sf q=m_*C_p*\Delta T\)

where q is the heat absorbed/released; the convention is positive when absorbed, negative when released.

m is the mass of the substance whose temp is being monitored

Cp is the specific heat capacity - sometimes this is of water alone, sometimes it's the whole calorimeter as a system.

\(\Delta T\) is the \(\sf change\) in temperature of the system, this can be expanded to \(\Delta T=T_f-T_i\)

Answer 3

The enthalpy of fusion is: \(\sf \Delta H_{fusion}=\dfrac{q}{mass}\)

Answer 4

Where do I put what in the equation?

Answer 5

Use the mass of water, the specific heat capacity of water (4.184 J/\(^oC\)g), and the initial and final temperatures of the water.

Answer 6

So does my equation need to look like this... q=457g(4.184)/*Cg(0) for the first trial?

Answer 7

nope. you need the mass of the water ONLY, you're including the mass of the calorimeter in the value you posted above.

You also need the change in temp, the water started at 20 celsius then decreased to 0 celsius.

Answer 8

OK so it needs to look like this then... q=48(4.184)/*Cg(20)

Answer 9

almost.

\(\Delta T =T_f-T_i=0~^oC-20~^oC=-20~^oC\)

the heat needs to be changed in sign after, so this step is redundant..

next use that second equation with the mass of the ICE (not the water)

Answer 10

q=48(4.184)/*Cg(-20) Change in fusion = q/41 (This should be the final right?)