A12 mm steel ball cools from1225 K to475 K in ambient325 K. Heat-transfer coefficient30 W/(m² K), density7800 kg/m³ and cp600 J/(kg K). Using lumped capacitance, find cooling time in seconds (one decimal).
The lumped relation is $$(T-T_\infty)/(T_i-T_\infty)=e^{-hAt/(\rho Vc_p)}.$$ A sphere has $V/A=d/6=0.002$ m, so the thermal time constant is $7800(600)(0.002)/30=312$ s. The temperature ratio is $(1225-325)/(475-325)=6$. Therefore $t=312\ln6=559.0$ s. The method assumes negligible internal temperature gradients.
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