How Does Doubling Time Affect Specific Heat Capacity in Thermal Physics?

[Doubell]

Homework Statement [/b]


E= 13600J, MAss of liquid = 0.1 kg, Temperature change = 25K
2. Homework Equations [/b]

Pt/m*delta T

The Attempt at a Solution

what effect would doubling the time have on the specific heat capacity of the liquid? what i did was to say if the time is doubled then the energy is doubled since the current * voltage = power and multiplying this by time = energy so if the time is doubled the energy is doubled. and if everything remains contant such as the temperature change and the mass. then a value for energy twice the initial value would be divided by the same value of mass and delta T. therefore the specific heat capacity would be doubled. is this justified? i would love to see a resolve to the problem.

 

[gneill]

You haven't actually stated the problem, the scenario, or a proper equation (which, generally speaking, should involve an equals sign).

 

[Doubell]

Homework Statement [/b]


E= 13600J, MAss of liquid = 0.1 kg, Temperature change = 25K
2. Homework Equations [/b]

Pt/m*delta T

The Attempt at a Solution

what effect would doubling the time have on the specific heat capacity of the liquid? what i did was to say if the time is doubled then the energy is doubled since the current * voltage = power and multiplying this by time = energy so if the time is doubled the energy is doubled. and if everything remains contant such as the temperature change and the mass. then a value for energy twice the initial value would be divided by the same value of mass and delta T. therefore the specific heat capacity would be doubled. is this justified? i would love to see a resolve to the problem.

The problem is based on the electrical method for finding specific heat capacity. and using the electrical method E=IVt = Pt. for finding specific heat capacity E = m * c * (T2-T1). therefore IVt = m*c*(T2-T1). hence c results to Ivt/m*(t2-t1). the solution i used as described above utilises this equation. and as i said if the time is doubled then E= IV(2t). this results in the value for energy becoming 2E. divivding 2E/m*(T2-T1) results in c being twice as large which supports my hypothesis that doubling the time results in the specific heat capacity being twice as large. just want to see if anyone agrees or disagree and hopefully show me another solution.

 

[gneill]

Why would you assume that (T2 - T1) remains the same? Wouldn't you expect the sample to get hotter if more heat is infused?

 

[Doubell]

well the question just asked what would happen to the specific heat capacity if the time was doubled. It didnt mention a temperature change so i had to assume temperatue change was kept constant.

 

[gneill]

well the question just asked what would happen to the specific heat capacity if the time was doubled. It didnt mention a temperature change so i had to assume temperatue change was kept constant.

I don't think that you can make that assumption. Heat capacities of materials don't change with time in general.