Question

5 g of ice of temp.20 Celsius in mixed in 5 g water at temp.30 Celsius

Answer 1

SUBJECT .

Physics. [ Thermal physics]

QUESTION TYPE.

when mix ice and water then find final temperature.

FORMULA USED.

$t = \frac{m(1)t(1) + m(2)t(2)}{m(1) + m(2)}$

t= Resultant temperature after mixing

m(1) = mass of water

t(1) = temperature of water

m(2)=mass of ice

t(2)=temperature of ice

FORMULA EXPLANATION.

LOOK AT ATTACHMENT .

SOLUTION .

GIVEN

m(1) = mass of water. =5g =5×1000=5000kg

t(1) = temperature of water

=30°C=30+273°=303K

m(2)=mass of ice=5g=5000kg

t(2)=temperature of water

=20°C=20°C+273=293K

USING FORMULA.

$t = \frac{m(1)t(1) + m(2)t(2)}{m(1) + m(2)}$

$t = = \frac{(5000 \times 303) + (5000 \times 293)kg. \: k}{(5000 + 5000)kg}$

$t = \frac{5000(293 + 303)k}{10000}$

$t = \frac{(293 + 303)k}{2}$

$= \frac{596}{2} k$

$= 444.5k$

T=444.5K

In °c

K=273+°C

=>

°C=K-273

so,

T(°C)=444.5-273

T(°C)=171.5°C

HENCE

Resultant temperature = 171.5°C. or 144.5 K