Question

Let the latent heat of fusion of ice be denoted by L, and let the specific heat of water be denoted by C. A mass m1 of ice at 0°C is mixed with mass m2 of liquid water at T°C, and all the ice melts. Assuming that no heat is gained from or lost to the surroundings, the final temperature is:

Answer 1

Answer:

ur ans is 72g

Explanation:

Let, mass of water used =m

Since, Heat gained = Heat lost

m

1

cΔt

1

=mL+m

2

cΔt

2

m×4.2×(60−10)=(40×336)+(40×4.2×(10−0))

m×4.2×50=(40×336)+1680

Therefore, the mass of water used, m=

42×5

13440+1680

=

210

15120

=72g

Answer 2

Answer:

Final Temperature $=\frac{m_{2}C- m_{1} L}{m_{1}C + m_{2}C }$

Explanation:

The question can be solved using the theorem of Conservation of Energy.

The energy released by the water in loosing its temperature is gained by ice to convert into liquid and raise its temperature.

Let the final temperature of the mixture be x then,

Heat released by water $Q_{water} =m_{2} C(T-x)$

Heat gained by water $Q_{ice}=m_{1}L +m_{1} C(x-0)$

Since the heat released by water is gained by ice,

                                    $Q_{water}= Q_{ice}$

                        $m_{2}C(T-x)=m_{1}L +m_{1} C(x-0)$

On solving for x,

                                     $x=\frac{m_{2}C- m_{1} L}{m_{1}C + m_{2}C }$

$Hence, Final\ temperature=\frac{m_{2}C- m_{1} L}{m_{1}C + m_{2}C }$