# Physics : How do I calculate percent uncertainty?

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I can however try to get more funds to tip well. Thanks and

sorry.

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

However im not sure these are right so I need help figuring out

how to correctly do this or for someone to check if it is

right?