Respuesta :
The heat transfer of the system is m . 4200 (35 - T) where m is mass flowrate and T is temperature.
The heat transfer to the system depends on the heat released from both gold and water. The total heat received by the system will equal to total heat released by objects. It should follow
Q released = Q received
The heat can be defined by
Q = m. c . ΔT
where Q is heat, m is flowrate of mass, c is the specific heat constant and ΔT is the change in temperature.
The given parameters are
T1 = 25⁰ C
T2 = 45⁰ C
T3 = 35⁰ C
A = 5mm x 20 mm = 100 mm²
c = 4200 J/kg⁰ C
Let's assume that the value of equilibrium temperature is T and the initial temperature is the same as the inlet water. Hence,
Q released = Q received
Q in + Qsurface - Qout = Qsys
m input . c . (T-T1) + H - m output . c . (T3 - T) = m system . c . (T-T1)
The flowrate is same. Thus, the heat transfer is
H = m.c ((T-T1) + (T3-T) - (T-T1)
H = m . 4200 (T-25) + (35 - T) - T+ 25)
H = m . 4200 (35 - T)
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