Question

$1\text{kg}$ of ice at $0^{\circ }\text{C}$ is mixed with $1\text{kg}$ of steam at ${100}^{\circ }\text{C}$. Find the mass of steam when the mixture has reached thermal equilibrium.
$[L_{fusion,ice}=3.36\times {10}^5\text{J/kg},$$L_{vaporization,water}=2.26\times {10}^6\text{J/kg},$$C_{water}=4200\text{J/kg-K}]$

• A. $800\text{g}$
• B. $665\text{g}$
• C. $500\text{g}$
• D. $400\text{g}$

Answer 1

The correct option is B $665\text{g}$

We know that in same amount of ice and steam, steam has more amount of heat energy. So. steam will melt the ice.

Heat that can be released by steam when convert to water$\left({100}^{\circ }\text{C}\right)$ -
$Q=mL$ $[$ Phase Change $]$
$=1\times 2.26\times {10}^6$ $[L_{vaporization,water}=2.26\times {10}^6\text{J/kg}]$
$=2.26\times {10}^6\text{J}$............(1)

Heat required by ice to melt into water$\left(0^{\circ }\text{C}\right)$ :-
$Q=mL$ $[$ Phase Change $]$
$=1\times 3.36\times {10}^5$ $[L_{fusion,ice}=3.36\times {10}^5\text{J/kg}]$
$=3.36\times {10}^5\text{J}$............(2)

Heat required by water$\left(0^{\circ }\text{C}\right)$ to raise its temperature to ${100}^{\circ }\text{C}$
$Q=ms\Delta T$ $[$ Temperature Change $]$
$=1\times 4200\times 100$ $[C_{water}=4200\text{J/kg-K}]]$
$=4.2\times {10}^5\text{J}$............(3)

Total heat required$=7.56\times {10}^5\text{J}$....(4)
$[$ from (2) and (3) $]$

As Heat required $<$ Heat released
$[$ from (1) and (4) $]$

Therefore, only some amount of steam will convert to water $\left({100}^{\circ }\text{C}\right)$ to provide required heat.

Let amount of steam converting to water be $x\text{kg}$
$\therefore$ Heat released $=mL=x\times 2.26\times {10}^6\text{J}$.......(5)

On comparing (4) and (5)
$x=0.335\text{kg}=335\text{g}$

Therefore, amount of steam left is $1000-335=665\text{g}$