# I had a lab today in class it

FratBro23
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Physics
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Question description

I had a lab today in class it was Latent Heat of Fusion.
My professor was rude and unclear about how to calculate the percent uncertainty or is it percent error??

The Equation Given was :
m_w C_w(T_h - T_F) + m_c C_c(T_h - T_F) = m_i C_i(0 - T_i) + (m_i* L_f) +m_i C_w(T_F-0)

To find Latent Heat of Fusion it is modified to
L_f = ( m_w C_w(T_h - T_F) + m_c C_c(T_h - T_F) - m_i C_i(0 - T_i) - m_i C_w(T_F-0) )/ m_i

Masses

mass of water (m_w) is 215.1 g +/- 0.1g
mass of Aluminum calorimeter (m_c) is 70.3g +/- 0.1g
mass of ice (m_i) is 8.9g +/- 0.1 g

Specific Heats or C's

C_c Aluminum 0.215 cal/(g*C)
C_w water 1.00 cal/(g*C)
C_i ice 0.5000 cal/(g*C)

Temps

T_H temp of hot water 48.1 +/- 0.1 C
T_I temp of Ice 0.1 +/- 0.1 C
T_f Final temp 43.2 +/- 0.1 C

m_wc_w(T_h-T_F) = 215.1g(1)(48.1-43.2)= 1053.99
m_cC_c(T-h-T_F) = 70.3g(0.215)(48.1-43.2)= 74.06
m_iC_i(0-Ti) = 8.9G(0.500)(0-0.0) = 0
m_i C_w(T_f-0)= 8.9g(1)(43.2)= 384.48

L_F = all sums added up or subtracted depending on the terms. divided by the mass of ice

L_f = 7346.57/8.9 = 83.547

so using all of this how do I calculate the percent error for the Latent Heat of fusion and for each of
parts shown above like 384.48 =+/- ???.

For L_f I got 83.547 +/- 6.36 cal/g
percent uncertainty or error?

m_wC_w or p1 .041

m_cC_C p2 .041

m_iC_i p4 0.0013

m_iC_w p3 .012
If some one good show me how to do this properly or show where I went wrong that would be awesome.

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