Yatayyat
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Been stuck on this question for ages!

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I'm not sure what formula to use here in order to solve this problem?

Any help would be appreciated! Thanks
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GgbroTG
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You need to use the specific heat formula,



Q = mc\Delta \theta

where Q is the (heat) energy into the system (out if negative), m is the mass of the material, c is its specific heat capacity and \theta is the temperature. Since you're looking for the rate of heat loss, you can introduce the rate of flow of mass to the right hand side of the equation to equalise.



P_Q = \frac{\Delta m}{\Delta t} c\Delta \theta

\\\

{\rm since} m = \rho V

\\\

P_Q = \rho \frac{\Delta V}{\Delta t} c\Delta \theta

where \frac{\Delta V}{\Delta t} is the rate of flow of volume and \rho is the density of water, as stated in the question.

Then substituting values in:



P_Q = 1000\cdot (1\times 10^{-5}) \cdot 4200 \cdot (31.5 - 73.7)

\\\

P_Q = -1772.4 {\rm W}

Therefore, the rate of heat loss is 1772.4 W. Technically speaking, we only know this answer correct to 1 significant figure (so 2000 W), due to the given rate of flow of volume, but I'll assume that isn't important.
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Yatayyat
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(Original post by GgbroTG)
You need to use the specific heat formula,



Q = mc\Delta \theta

where Q is the (heat) energy into the system (out if negative), m is the mass of the material, c is its specific heat capacity and \theta is the temperature. Since you're looking for the rate of heat loss, you can introduce the rate of flow of mass to the right hand side of the equation to equalise.



P_Q = \frac{\Delta m}{\Delta t} c\Delta \theta

\\\

{\rm since} m = \rho V

\\\

P_Q = \rho \frac{\Delta V}{\Delta t} c\Delta \theta

where \frac{\Delta V}{\Delta t} is the rate of flow of volume and \rho is the density of water, as stated in the question.

Then substituting values in:



P_Q = 1000\cdot (1\times 10^{-5}) \cdot 4200 \cdot (31.5 - 73.7)

\\\

P_Q = -1772.4 {\rm W}

Therefore, the rate of heat loss is 1772.4 W. Technically speaking, we only know this answer correct to 1 significant figure (so 2000 W), due to the given rate of flow of volume, but I'll assume that isn't important.
Thanks this was helpful! I understand now
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localmemelord
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#4
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(Original post by GgbroTG)
You need to use the specific heat formula,



Q = mc\Delta \theta

where Q is the (heat) energy into the system (out if negative), m is the mass of the material, c is its specific heat capacity and \theta is the temperature. Since you're looking for the rate of heat loss, you can introduce the rate of flow of mass to the right hand side of the equation to equalise.



P_Q = \frac{\Delta m}{\Delta t} c\Delta \theta

\\\

{\rm since} m = \rho V

\\\

P_Q = \rho \frac{\Delta V}{\Delta t} c\Delta \theta

where \frac{\Delta V}{\Delta t} is the rate of flow of volume and \rho is the density of water, as stated in the question.

Then substituting values in:



P_Q = 1000\cdot (1\times 10^{-5}) \cdot 4200 \cdot (31.5 - 73.7)

\\\

P_Q = -1772.4 {\rm W}

Therefore, the rate of heat loss is 1772.4 W. Technically speaking, we only know this answer correct to 1 significant figure (so 2000 W), due to the given rate of flow of volume, but I'll assume that isn't important.
not trying to be a bummer or anything like that but you shouldve given hits or tips to the person asking for help and not given the answer to them straight away.
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GgbroTG
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(Original post by localmemelord)
not trying to be a bummer or anything like that but you shouldve given hits or tips to the person asking for help and not given the answer to them straight away.
That's fair; I did consider leaving it at just specific heat capacity, but sometimes I feel a worked solution is more useful, especially when the question is a little more unconventional. Granted, I could've at least left the answer itself out.
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Eimmanuel
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(Original post by GgbroTG)
That's fair; I did consider leaving it at just specific heat capacity, but sometimes I feel a worked solution is more useful, especially when the question is a little more unconventional. Granted, I could've at least left the answer itself out.
You can consider using the spoiler alert function to hide your solution/answer in future.
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GgbroTG
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(Original post by Eimmanuel)
You can consider using the spoiler alert function to hide your solution/answer in future.
A very good suggestion; I'll definitely consider doing so.
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