Before you place your bid in. I can only pay in the range I put. 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) = 0m_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 = 83.547so 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/gpercent uncertainty or error?m_wC_w or p1 .041m_cC_C p2 .041m_iC_i p4 0.0013m_iC_w p3 .012However 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?
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