﻿ Basic Heat Energy Transfer

Basic Heat energy Transfer

Strings (SiPjAjk) = S7P5A51     Base Sequence = 12735     String Sequence = 12735 - 5 - 51

Expressions Of Pj Problems
Basic Heat Energy Transfer
Math A piece of Aluminum of mass 3.90 g at temperature 99.3oC is immersed into a 10.0 cm3 of water at 22.6 oC (figure 7.2). Determine the final temperature of the system.

The strings: S7P5A51 (Change - Physical).

The math:
Pj Problem of Interest is of type Change (physical). Assumption: system is insulated
Let q = heat gained or lost; m = mass in grams; Cp = specific heat;
ΔT = change in temperature = Tfinal - Tinitial (when heat is gained).
ΔT = change in temperature = Tinitial - Tfinal (when heat is lost).
By the law of conservation of energy:
q = m(ΔT)Cp --------(1)
qlost = qgained -------(2)

Now mass of water = Density x Volume = 1 x 10 g/cm3.
So from equation (2):
(3.90)(99.3 - Tfinal)(0.903) = (10)(22.6 - Tfinal)(4.18)
So, final temperature of system, Tfinal = 28.6oC. The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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