phy

1. 1.) An elastic titanium peg was being led into the broken ulna of the
lower arm. Following the environmental temperature, the
peripheral body temperature can change even by 10°C. By how
many millimeters would this change the length of a 28 cm long
titan peg?
The linear thermal expansion coefficient of titan is 8.64· 10¯⁶ 1/°C.
Data: ΔT = 10°C
Ɩ0 = 28 cm
α = 8,64· 10¯⁶ 1/°C
Relationship
Substitution
ΔƖ = Ɩ0 ∙α ∙ ΔT
Question: ΔƖ= ? (mm)
ΔƖ = 28 cm ∙ 8,64· 10¯⁶ 1/°C ∙ 10°C = 2419,2· 10¯⁶ cm = 0,024 mm
T0
T0 + ΔT
Δl
l0

2. 2.) The diameter of a titanium made and approximately spherical
femurhead is 2.7 cm. How large change of temperature would
increase its volume by 0.1%? The linear thermal expansion
coefficient of titan is 8.64· 10¯⁶ 1/°C.
Data: d = 2,7 cm
ΔV = V0· 0,001
α = 8,64· 10¯⁶ 1/°C
Question: ΔT = ? °C
Relationship
Substitution
ΔV = V0 ∙3α ∙ ΔT = V0· 0,001
V0 ∙3α ∙ ΔT = V0· 0,001
3α ∙ ΔT = 0,001
ΔT =
0,001
3α
=
0,001
3·8,64· 10¯⁶ 1/°C
= 38,6 °C
𝐕𝟎

3. 3.) .) By how many °C will the body temperature of a 14.5-kg child
with fever decrease if the heat transferred to the cooling bath is 70
kJ? (The specific heat of the human body is 4190 J/(kg·°C), the heat
production of the body is neglected.)
Data: m = 14,5 kg
ΔQ = 70 kJ
c = 4190 J/(kg·°C) = 4190
J
kg·°C
Question: ΔT = ? °C
Relationship:
So:
Sustitution:
ΔQ
c∙m
= ΔT
ΔQ = c∙m∙ ΔT
ΔT =
70∙103 J
4190 J
kg·°C
∙14,5 kg
= 1,15°C

4. 4.) In a summer training, the body of a 80-kg athlete evaporates
1.13 litres of water. How many kilojoules of heat was withdrawn
thereby? (The heat of evaporation for water is 2.25·10⁶ J/kg.)
Data: m = 80 kg
ΔV = 1,13 l (mwat = 1,13 kg)
Lp = 2,25·10⁶ J/kg
Question: ΔQ =? kJ
Relationship
Substitution
ΔQ = Lp ·mwat
ΔQ = 2,25·10⁶ J/kg · 1,13 kg = 2542,5 kJ

5. 5.) How much heat is being lost because of heat conduction in every
second, if someone covers the whole surface of their body tightly
with a 0.5 cm thick layer of wool? (The temperature of skin and
environment are 35°C and 20 °C, respectively, the area of body
surface is 1.5 m², the thermal conductivity of wool is 0.04
W/(m·K).)
Data: : t = 1s
x = 0,5 cm = 0,005 m
Tbőr= 35 °C
Tkör= 20 °C
A = 1,5 m²
k = 0,04 W/(m·K)
Question: Q = ? J
Relationship
Substitution
Q
t
= -k·A
T2 − T1
x
Q =1s·0,04 W/(m·K) · 1,5 m²
−(20°C−35 °C )
0,005 m
= 1·0,04·1,5
35 −20
0,5
·
s·W·m²·K
m·K·m
Q = 180 J
skin air
wool
∆Q/∆t
d

6. 6.) The forensic medical expert measures the temperature of a dead
body found in a 20°C warm room 27°C. How many hours before did
the murder happen, assuming that the temperature of the victim was
36.5°C at the time of the murder? The cooling rate coefficient of
human body is 4·10¯⁵ 1/s.
Data: Tenv= 20 °C
T2= 27 °C
T1= 36,5 °C
k = 4·10¯⁵ 1/s.
Question: t = ? h Relationship
Substitution
T2 − Tenv
T1 − Tenv
= e-kt
T2 = Tenv + (T1 – Tenv)·e-kt
ln
T2 − Tenv
T1 −Tenv
= -k·t
ln
27 °C−20 °C
36,5 °C−20 °C
= - 4·10¯⁵ 1/s ·t
-0,857 = - 4·10¯⁵ 1/s ·t
21436,26 s = t
5,95 h = t