Another calorimeter problem, and %error

[erdfcvtyghbn]

Homework Statement

calorimeter mass = 70 g
specific heat of calorimeter = .1 cal/gC
cal. and water mass = 200g
temp of water and cal = 65 C
mass of cal, water, and ice = 220 g
temp of ice = 2 C
final temp of cal, water, and ice = 30 C
Find Lf for this ice.
Find the percent error.

Homework Equations

m Lf + m c (change in temp) = m c (change in temp) + m c (change in temp)
left of equals sign is for ice, 1st m c T on right is water, 2nd is calorimeter

The Attempt at a Solution

20 Lf + 20 2.09 28 = 130 4.19 -35 + 70 (.1 x 4.19) -35
i got -1063.07 for Lf, but I am not sure if its right, and I am not sure how to find percent error

 

[Mindscrape]

The percent error will be the discrepancy between the experimental value and the known value over the known value.
$$percenterror = \frac{|x_{exp} - x_{known}|}{x_{known}}*100$$

 

[m2003]

I would first commend the student for their attempt at solving the problem and using the appropriate equations. However, I would also provide some guidance and clarification to ensure accuracy in their calculations.

To find the latent heat of fusion (Lf) for the ice, the equation should be set up as follows:

miceLf + micec(deltaT) = mwaterc(deltaT) + mcalcc(deltaT)

where mice is the mass of the ice, c is the specific heat of water (4.19 cal/gC), and deltaT is the change in temperature (65-2 = 63 C).

Substituting the given values, we get:

(20g)Lf + (20g)(1 cal/gC)(63C) = (130g)(4.19 cal/gC)(-35C) + (70g)(0.1 cal/gC)(-35C)

Simplifying, we get:

20Lf + 1260 = -1617.5 + (-245)

20Lf = -1122.5

Lf = -1122.5/20 = -56.125 cal/g

Note that the negative sign indicates that heat is being removed from the system, which is expected since the ice is melting.

To find the percent error, we need to compare our calculated value of Lf with the accepted value of Lf for water, which is 80 cal/g. The percent error is calculated as:

Percent error = (|calculated value - accepted value|/accepted value) x 100%

Substituting our values, we get:

Percent error = (|-56.125 - 80|/80) x 100% = 44.84375%

This means that our calculated value deviates from the accepted value by approximately 45%. This could be due to experimental errors or inaccuracies in our measurements. I would suggest repeating the experiment and taking more precise measurements to reduce the percent error.