p6.4
Specific heat of copper, C = 0.0932 cal / ( g.Co
)
Mass, m = 750 g
Raise of temperature, ΔT = 40 - 10 =
30Co
Heat absorbed, Q = m C ΔT
=
750 * 0.0932 * 30
=
2097 cal
=
2097 * 4.186 J
=
8778
J
p6.4 and 0 °C respectively. Determine the steady state heat rate through this rod via conduction....
need help with c and d
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