Heat transfer equation (temperature difference)

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Homework Help Overview

The discussion revolves around the heat transfer equation mcΔT, specifically focusing on the interpretation of the temperature difference ΔT in various contexts, such as heat loss and heat gain. Participants are examining the conditions under which ΔT should be expressed as Tf - Ti versus Ti - Tf.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the implications of using different expressions for ΔT in the context of heat transfer, questioning when to apply each form. There is discussion about the signs associated with heat gained and lost, and how these relate to the temperature differences.

Discussion Status

The conversation is ongoing, with participants providing insights into the relationship between heat transfer and temperature differences. Some have suggested that the negative sign in heat loss calculations is necessary, while others are clarifying how the expressions relate to energy gained or lost. There appears to be a productive exploration of the concepts without a clear consensus yet.

Contextual Notes

Participants are considering the implications of sign conventions in the context of energy transfer, particularly in scenarios involving heat loss and gain. There is an acknowledgment of the need to maintain consistency in the application of the heat transfer equation.

MechaMZ
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Homework Statement



For the equation mc\DeltaT, i suppose \DeltaT should be Tf - Ti.

However, at what condition we should let \DeltaT be Ti - Tf?

I did come across a question which requires \DeltaT be Ti - Tf for a heat lost by an object.

eg,
heat lost by copper = heat absorbed by beaker & water
mc(Ti - Tf) = mbeakercbeaker(Tf - Ti) + mwatercwater(Tf - Ti)

I think perhaps we need to let mc\DeltaT as a positive value since it just a magnitude? however, the value is not the same(pretty close) if i let it be Ti - Tf.

Need help, thanks =)
 
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anyone knows?
 
mc\Delta T=mc(T_f-T_i) is always the energy gained. The energy lost by the copper was then -mc\Delta T, which is equivalent to the other expression.
 
however, for the example above, why the initial temp and final temp is reversed?
 
Because -mc(T_f-T_i)=mc(T_i-T_f). Is this what you mean?
 
so we should write heat lost by copper as negative since the heat is released out, but the result of -mc(T_f-T_i) will still be positive right?
 
If you're saying that we should include the minus sign prefactor because we're considering the amount of energy leaving, but that the sign of -mc(T_f-T_i) is positive because T_f<T_i, I agree.
 

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