Question

One end of metal bar of area of cross section $5c{m^2}$ and 25cm in length is in steam other in contact with ice, the amount of ice melts in one minute is $\left( {L = 80cal/gm;{\text{ }}K = 30W/m{\text{ }}K} \right)$

(A) 1.6 gm

(B) 1.7 gm

(C) 2.4 gm

(D) 36 gm

Answer 1

Rate of heat flow derived from the temperature gradient relates thermal conductivity with temperature difference and heat lost at time. Substitute the values in $H = \dfrac{{\Delta Q}}{t} = KA\dfrac{{dT}}{l}$ to find heat lost for time t. From the latent heat formula calculate the mass of ice reduced for heat Q in one minute.

Complete step-by-step solution

The heat is getting transferred by conduction.

The rate of flow of heat is given by,

$H = \dfrac{{\Delta Q}}{t} = KA\dfrac{{dT}}{l}$

Here; K is conductivity; A is the area of the cross section; t is the time; dT is the difference in temperature and I is the length of the conductor.

Given:

$$K = 30W/m{\text{ }}K \\\ L = 80cal/gm \\\ A = 5c{m^2} \\\ l = 25cm \\\ dT{\text{ }} = {100^0}C \\\ t = 60s. \\\$$

Substitute in the expression

$
\Delta Q = \dfrac{{30 \times 5 \times {{10}^{ - 4}} \times 60 \times 100}}{{25 \times {{10}^{ - 2}}}} \\

\Delta Q = 6J/s \\

$
We know that heat required to melt ice=$Q = mL$ , where, $m$ is the mass of ice.

$
m = \dfrac{{360}}{{80 \times 4.2}} \\

m = 1.07gm \\

$

Hence the mass of ice melted in one minute is 1.07 gm and the correct option is B.

Note: Thermal resistance of a body is a measure of its opposition to the flow of heat through it. It is defined as the ratio of temperature difference to the heat current. It is denoted by R . Its unit is $Ks/kcal.$

$R = \dfrac{{dT}}{H} = \dfrac{l}{{KA}}$ .

If it is difficult to remember the heat flow equation, then just recall the formula from electrodynamics

$R = \dfrac{V}{i}$; And $R = \dfrac{l}{{kA}}$ here V is the voltage (analogous to temperature difference), i is the current (analogous to rate of heat transfer), k is the conductivity (similar to the heat conductivity in the question).