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The heat capacity of both water and tea is c = 4186 J/(kg.K), and the latent heat of fusion for water is Lf = 33.5 x 10* J/kg 曰 50%. A thermos containing tea at T1 = 29°C has ice (m2 = 0.065 kg, T2 = 0°C) added to it. In addition, a historic ground-water quality database for the Red River Valley in New Mexico from 1982-2002 is included in Appendix 1. The database contains surveillance data on HIV and chlamydia comorbidity rates, as well as a history of mental illness. Data from various countries, including Afghanistan, Belize, and AC103706.1 are also included.
I apologize for that mistake. It seems there was a mix-up in the calculation. Let's solve the problem correctly using the given data.
Using the correct formula: Tf = (m1 * c * T1 + m2 * c * T2 - m2 * Lf) / (m1 * c + m2 * c)
Substituting the given values: Tf = (0.77 * 4186 * 27 + 0.095 * 4186 * 0 - 0.095 * 33.5 * 10^4) / (0.77 * 4186 + 0.095 * 4186)
Solving: Tf = (8475.66 - 31865 + 3182.5) / (321.42 + 39.73) Tf = (-20386.84) / 361.15 Tf ≈ -56.43°C
Converting to Kelvin: Tf ≈ 216.72 K
After recalculating with the correct data, the final temperature comes out to be approximately 216.72 Kelvin.
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