An ice cube of dimensions 60cm × 50cm × 20cm is placed in an...

Using the equation of conduction,

$\dfrac{dQ}{dt} = \dfrac{KA\DeltaT}{l}$

Now the total area of the box will be,

$A = 2\left(0 . 6 \times 0 . 5 + 0 . 5 \times 0 . 2 + 0 . 2 \times 0 . 6\right)$

$A = 2\left(0 . 3 + 0 . 1 + 0 . 12\right)$

$A = 2\left(0 . 52\right) = 1 . 04{m}^{2}$

Therefore,

$\dfrac{dQ}{dt} = \dfrac{KA\DeltaT}{l} = \dfrac{0 . 05 \times 1 . 04 \times 40}{0 . 01} = 208J{s}^{- 1}$

This heat will melt the ice. Therefore,

$\dfrac{dQ}{dt} = \dfrac{d m}{d t}L ⇒ 208 = \dfrac{d m}{d t} \times 3 . 4 \times {10}^{5} ⇒ \dfrac{d m}{d t} = 61 \times {10}^{- 5}kg{s}^{- 1}$