17. Assignments thermoregulation (15)

Question 1

ME scheme: what does it look like? For thermoregulation assignments

Answer 1

ME

-> maintenance (heat 100%)
-> Production (product X%, Heat 100-X%)
-> Work (Power 25%, heat 75%)

Question 2

Pig = 80 kg
Calculate total heat load of this pig, given..

• met weight = 80^0.75 = 26.7
• feed intake = 2.5 ME
• Production efficiency = 70%

Q = What is the total heat load?

Answer 2

BMR = brody formula = 400*80^0,75 = 10,680 kj/day = MEm

Food intake = 2.5ME = 2.510,680 = 26,700 kj/day

Work is not mentioned. Maintenance is 1*ME
Production is therefore 2.5-1 = 1.5x ME
Product efficiency = 70%, heat load is therefore 30%.

1.5 * 10,680 = 16020 kj. 30% = 4806 kj
+ 1*10,680 = 15486 kJ

Question 3

Pig = 80 kg
Calculate total heat load of this pig, given..

• met weight = 80^0.75 = 26.7
• feed intake = 2.5 ME
• Production efficiency = 70%

Q = What is the composition of the heat, product and feed load?

Answer 3

Maintenance heat = 1ME
Production heat = 0.3 1.5 = 0,45ME
(total = 1,45)
Product = 0.7 1.5 = 1.05ME
Feed = 2.5ME

Question 4

Pig = 80 kg
Calculate total heat load of this pig, given..

• met weight = 80^0.75 = 26.7
• feed intake = 2.5 ME
• Production efficiency = 70%

Q = How much O2 is consumed by the pig?

Answer 4

Thumb rule: 1 L O2 = 20 kJ/L
Oxygen is not needed to store nutrients. Product is not used for oxygen consumption.
You use the value of heat to calculate the total amount of oxygen consumed.

Energy of heat = 15,486 (kJ/day)/20(kJ/L) = 744 liter O2/day

Question 5

1 kcal = … joules

Answer 5

4200

Question 6

What is a formula to calculate the time (min) required for a body to increase its temperature with 1*C?

Answer 6

time (min) = 15*W^0.25 (weight in kg)

E.g. One-day chicken of 50 gr:
0.05 kg = 15*0.05^0.25 = 7.1 min

Question 7

The larger you are, the less/more time you relatively need to warm up because …

Answer 7

you have way more heat production from the volume, compared to loss which is related to the surface area.

Question 8

Available feed = 2ME.
What is the maximal energetic efficiency for production? When is it maximal? Use heat diagram p. 164

Answer 8

1ME for maintenance
Therefore 1ME for product
0.6ME = 60% for product,
04ME = for product heat + 1ME = 1.4ME for total heat.

Question 9

Use heat diagram p. 164: what is the LCT for maintenance?

Answer 9

20*C

Question 10

Is Thermoneutral zone fixed for a certain level of heat production (no species differences!)? True/false

Answer 10

False

Question 11

Thermoneutral zone does not rely on environmental factors: true/false

Answer 11

False

Question 12

Heat production is constant throughout the thermoneutral zone. True/false

Answer 12

True

Question 13

Heat loss is constant throughout thermoneutral zone: true/false

Answer 13

False (once you pass your body temperature, besides your normal standard heat production, there is more heat load from the environment. Total heat load is more than you produce by yourself. Heat loss also needs to increase, by evaporation)

Question 14

Is Heat production the same at all LCT’s?

Answer 14

No

Question 15

 Is Heat loss the same at all UCT’s?

Answer 15

YES
All UCP’s are on this line (B0 – Bx): on every UCP there is the same heat loss.
Heat loss at UCP is always at maximum evaporation capacity

Question 16

Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O

1. Calculate heat production (times MEm)

Answer 16

Power = 14 kj/min = 25% of 2ME (of 14)
100% = 144 = 56 kj/min
Maintenance = 7 kj/min = 1*ME
Nothing is said about production, therefore we do not take this into account.
7 + 75% * 56 = 49 kj/min

Question 17

Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O

2. Calculate energy expenditure

Answer 17

total EE = 56 + 7 = 63 kJ/min

Question 18

Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O

3. Calculate max capacity for evaporation

Answer 18

heat of evap = 2.5 kj/gr H20
Max evap = 22.4 ml/min
2.5 * 22.4 = 56 kj/min

Question 19

Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O

4. Determine (graphically) max Ta that allows long lasting activity

Answer 19

Look at all max values, look at the MEm the cycler is at and determine temperature when looking at the UCT line.

Question 20

Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O

5. Determine external heat load

Answer 20

At cycling, look at total heat produced by cycler (7MEm) and max evaporative capacity (8ME)
8-1 = 1x MEm is external heat load (7kJ/min)

Question 21

What can happen to the chemical (gross) energy of nutrients after ingestion?

Answer 21

Loss in faeces, ME anabolic + catabolic processes -> ATP, excretion of N via urea in urine

Question 22

• What can happen to the energy trapped in ATP?

Answer 22

ATP: anabolic: storage, synthesis, heat, exercise

Question 23

Which metabolic processes may contribute to extrta heat production at low environmental temperatures?

Answer 23

• Shivering thermogenesis: muscle contraction
• Non-shivering thermogenesis:
  Metabolic basis (UCP: uncoupling proteins) and
  Substrate cycling and oxidative processes (change of substrate to have higher release of heat)

Question 24

What is the heat capacity of the human body?

Answer 24

4.2 J/g/*C

Question 25

RQ (CO2/O2) of normal diet = ?

Answer 25

0.8 (protein is 0.82)