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Compute Body Temperature Increase

GIVEN DATA

Jogger's heat energy generated after running half an hour Joules, J. Click the link for heat energy to calculate the equivalent calorie .

"9e5" means exponent 5, this is the input format for big numbers accepted by JavaScript number format use in most of the web browser. Mathematically written for manual calculation as 9 x 10⁵ or 9x10^5, but JavaScript number format used in most web browser don't accept 9 x 10⁵ or 9x10^5 as big number input.You can try it here and the answer is 0 or NaN (Not A Number), meaning the input is not recognize by computer as a number.

Jogger's body mass, m (convert pound mass lbm to kg) , kg

Use average specific heat of human body Joules (J) / kg * ° C

ANSWER: Jogger's body temperature increase = ° C

Solve for jogger's body temperature increase

Memory recall formula for heat energy removed or supplied (generated)

Heat Energy generated ( Q ) = mass of body in grams (m) * specific heat of body (c) * change in temperature ( Δ T)

Q = m * c * Δ T the unit of energy is in Joules (J)

By algebraic operation, you get the equation on how to calculate the change in temperature ( Δ T)

Δ T = Q / ( m * c )

= / ( * )

= ° C

New scenario calculator, compute heat generated energy Q in Joules Q = m * c * Δ T Answer Joules , heat generated energy

After memory recall of Heat Energy Formula, Q, next enter "NEW" data for what if scenario analysis, answer will be in blue button shown below. Remember do not enter comma ,
Input Box 1 Given heat energy in Joules (39372) , Q ,

Q = 900,000 joules can be written as 9e5 for human body jogging for half hour.

input box 2. m, Given mass of body in grams (1200), Input Box 2
For human body in kilogram mass, multiply the given kg. by 1000 to get the gram equivalent, for example 50 kg x 1000 = 50000 grams

Input Box 3 c, Given specific heat of body, for example copper alloy (0.386)
in Joule / gram ° C ,



Input Box 4 Δ T, Given delta or change in temperature of body
from 20 ° C to 105 ° C = 85 ,

Compute mass of the body of copper alloy in grams, Given Information Δ T = 85 (enter in input box 4) Heat energy, Q = 175000 Joules (enter in input box 1). You need to have list of known specific heat value of material, for example copper alloy,
c = 0.386 in Joule / gram ° C (enter in input box 3)

m = Q / c * Δ T Answer grams

Compute the change (delta) of temperature in degree ° C Given information
mass of the body, m = 5333.74 g (enter in input box 2) Heat energy absorbed, Q = 175000 Joules (enter in input box 1)

specific heat of copper alloy, c = 0.386 in Joule / gram ° C (enter in input box 3)

Assuming your material is still copper alloy and the mass is still the same (5,333.74 g) and the heat energy absorb by the copper alloy is Q = 175000 Joules. You will expect that the delta of temperature Δ T, will also increase. What is the value of this delta temperature in ° C ? Enter the mass (5,333.74) in Input Box 2

Δ T = Q / m * c Answer in ° C

If your material was changed, then find from the engineering reference table shown below the corresponding specific heat energy of the new material and enter on Box 3.

Compute the specific heat of the body in Joule / gram Δ ° CGiven Information
Heat energy absorbed, Q = 175000 Joules (enter in input box 1)

mass of the body, m = 5333.74 g (enter in input box 2) Change in temperature Δ T = 85 (enter in input box 4)

Specific heat of the material in question c = Q / m * Δ T

Answer Joule / gram Δ ° C , By knowing the specific heat generated or absorbed by a material, the material can be identified.

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