HP 48GII

Equation Reference 5-29
Heat Capacity (6, 1)
Equations:
QmcT∆⋅⋅=QmcTfTi–()⋅⋅=
Example:
Given: ∆T=15_°C, Ti=0_°C, m=10_kg, Q=25_kJ.
Solution: Tf=15_°C, c=.1667_kJ/(kg
∗ K)
Thermal Expansion (6, 2)
Equations:
Example:
Given: ∆T=15_°C, L=10_m, Tf=25_°C,
δ=1_cm.
Solution: Ti=10_°C,
α =6.6667E–5_1/°C.
Conduction (6, 3)
Equations:
q
k
A⋅
L
-----------
T∆⋅=q
kA⋅
L
-----------
Th Tc–()⋅=
Example:
Given: Tc=25_°C, Th=75_°C, A=12.5_m^2, L=1.5_cm, k=0.12_W/ (m
∗ K)
Solution: q=5000_W, ∆T= 50_°C.
δ
α
L T
∆ ⋅ ⋅
=
δ
α
L T
f
Ti
–
( )⋅ ⋅
=