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If internal temperature gradients are small, a body’s temperature during convective cooling follows an exponential response.
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Formula
Temperatures in K; h in W/(m2·K); A in m2; ρ in kg/m3; V in m3; cp in J/(kg·K); t in s
| Symbol | Meaning | Unit |
|---|---|---|
| Body temperature at time t | K | |
| T∞ | Ambient temperature | K |
| Initial body temperature at t=0 | K | |
| Convective conductance at the surface | W/K | |
| Total body heat capacity in the lumped model | J/K |
\frac{T(t)-T_\infty}{T_i-T_\infty}=e^{-hAt/(\rho V c_p)}The model assumes a spatially uniform body temperature, often justified by a small Biot number. Its time constant τ=ρVcp/(hA) combines thermal storage and surface transfer. It also assumes constant ambient temperature and h, with negligible internal gradients. For larger Biot numbers, solve a spatially resolved transient conduction problem.
The model assumes a spatially uniform body temperature, often justified by a small Biot number. Its time constant τ=ρVcp/(hA) combines thermal storage and surface transfer. It also assumes constant ambient temperature and h, with negligible internal gradients. For larger Biot numbers, solve a spatially resolved transient conduction problem.
If internal temperature gradients are small, a body’s temperature during convective cooling follows an exponential response.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Lumped-capacitance model of body temperature” and check its unit and magnitude.
A body starts 60 K above ambient and has τ=200 s.
After 200 s, the excess is 60/e≈22.1 K.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: Temperatures in K; h in W/(m2·K); A in m2; ρ in kg/m3; V in m3; cp in J/(kg·K); t in s.
Formula cluster
If internal temperature gradients are small, a body’s temperature during convective cooling follows an exponential response.
A body starts 60 K above ambient and has τ=200 s. After 200 s, the excess is 60/e≈22.1 K.
Estimate heating and cooling times for compact bodies with small Biot number.
Here is how to work through a typical Lumped-capacitance model of body temperature ((T−T∞)/(Ti−T∞) = exp[−hAt/(ρVcp)]) task step by step:
Task
Work through the numerical example for “Lumped-capacitance model of body temperature” and check its unit and magnitude.
Solution path
A body starts 60 K above ambient and has τ=200 s.
After 200 s, the excess is 60/e≈22.1 K.
τ=ρVcp/(hA) indicates how quickly the temperature difference from ambient decays. After one time constant, e−1≈36.8% of the initial difference remains, meaning about 63.2% has decayed. Greater heat capacity lengthens τ; more area or stronger convection shortens it. The single-temperature model assumes small internal gradients, often checked through Biot number. Its exponential response assumes constant properties, ambient conditions and a simple surface transfer. The time constant is heat storage divided by convection conductance; real systems may need temperature-dependent h and cp.
It is suitable when internal temperature equalizes quickly relative to surface transfer. Biot number is a common preliminary check; small values support the assumption. Large internal gradients require a spatial transient-conduction solution. Changes in h, cp or ambient conditions can also disturb the simple exponential curve. The single-temperature model assumes small internal gradients, often checked through Biot number. Its exponential response assumes constant properties, ambient conditions and a simple surface transfer. The time constant is heat storage divided by convection conductance; real systems may need temperature-dependent h and cp.
The excess above ambient is multiplied by e−1, so about 36.8% of the initial difference remains. From an initial 60 K excess, that is about 22.1 K. Body temperature is ambient plus this remaining difference, not 36.8% of its absolute Kelvin temperature. The single-temperature model assumes small internal gradients, often checked through Biot number. Its exponential response assumes constant properties, ambient conditions and a simple surface transfer. The time constant is heat storage divided by convection conductance; real systems may need temperature-dependent h and cp.
Write the quantity and unit from the variable list next to every value: Temperatures in K; h in W/(m2·K); A in m2; ρ in kg/m3; V in m3; cp in J/(kg·K); t in s. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example A body starts 60 K above ambient and has τ=200 s. After 200 s, the excess is 60/e≈22.1 K. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. The single-temperature model assumes small internal gradients, often checked through Biot number. Its exponential response assumes constant properties, ambient conditions and a simple surface transfer. The time constant is heat storage divided by convection conductance; real systems may need temperature-dependent h and cp.
The background states this starting condition: The model assumes a spatially uniform body temperature, often justified by a small Biot number. Its time constant. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. The single-temperature model assumes small internal gradients, often checked through Biot number. Its exponential response assumes constant properties, ambient conditions and a simple surface transfer. The time constant is heat storage divided by convection conductance; real systems may need temperature-dependent h and cp.
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Which formula describes Lumped-capacitance model of body temperature?
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How do you rearrange Lumped-capacitance model of body temperature for Base equation for the target quantity?
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Which common mistake happens with Lumped-capacitance model of body temperature?
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Create a curated FSRS exam set for (T−T∞)/(Ti−T∞) = exp[−hAt/(ρVcp)]: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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