Question

when 150gm ice of 0°c is mixed with 300gm water of 50°c then what is the resultant temperature of the mixture??​

Answer 1

why don't you check the textbook solved examples.

Answer 2

Let's See What Had Happened When 150gm of Ice is Mixed With 300 gm Water!

Firstly It Will Get Converted into Water and Then Both Will Attain a Certain Temperature!

So, Let's Se What we Have

Mass of Ice = 150 gm

Mass of Water = 300 gm

Specific heat of Water in Calories = 1 Cal.

Latent Heat (Ice → Water) = 80 Cal.

So, We Know

Heat Gained = Heat Lost

Let's say They attain a Stable Temperature T

So,

mL + mSΔT = MSΔT$_{1}$

[Remember Both are Having The Same Specific Heat Because Ice is Now Converted into Water]

Putting The Values

150*80 + 150*1*(T-0) = 300*1*(50-T)

80 + T = 2(50-T)

80 + T = 100 - 2T

3T = 20

T = 20/3

or

T = 6.67°

.

.

$\huge \blue{THANKS}$