Question

estimate the rate at which ice would melt in a wooden box 2.5 cm thick of inside measurement 100cmx60cmx40cm, assuming that the external temperature is 35°C and thermal conductivity of wood 0.169 W/mK

Answer 1

we know, $\frac{dQ}{dt}=kA\frac{dT}{dx}$

where k is thermal conductivity, A is cross sectional area , dT is change in temperature and dx is thickness.

here, k = 0.169w/mK

dT = 35°C = 273 + 35 = 308K

combined area of six faces = 2 x (1 x 0.6 + 0.6 x 0.4 + 0.4 x 1) m² = 2.48 m²

dx = 2.5cm = 0.025m

now, dQ/dt = 0.169 × 2.48 × 308/0.025

= 5,163.5584 J/s or, 1,229.41867Cal/s

hence, rate of melting of ice is 1,229.41867 Cal/s

Answer 2

Answer:

why ice is keept in wooden box