Question

300g of water at 25°c is added to 100g of ice at 0°c. the final temperature of the mixture is

Answer 1

Answer : The final temperature of the mixture is, 294.40 K

Solution :

$Q_{absorbed}=Q_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_{water}\times (T_{final}-T_1)=-[m_2\times c_{ice}\times (T_{final}-T_2)]$          .................(1)

where,

$m_1$ = mass of water = 300 g  = 0.3 kg     (1 kg = 1000 g)

$m_2$ = mass of ice = 100 g  = 0.1 kg

$T_{final}$ = final temperature

$T_1$ = temperature of water = $25^oC=273+25=298K$

$T_2$ = temperature of ice = $0^oC=273+0=273K$

c = specific heat of water = 4185.5 J/kg K

c = specific heat of ice = 2108 J/kg K

Now put all the given values in equation (1), we get

$(0.3kg)\times (4185.5J/kgK)\times (T_{final}-298K)=-[0.1kg\times (2108J/kgK)\times (T_{final}-273K)]$

$T_{final}=294.40K$

Therefore, the final temperature of the mixture is, 294.40 K

Answer 2

Qs:-300 of water at 25°c is added to 100 of ice at 0°c.The final temperature of the mixture is ?

Ans:- Hot body(Water)

mass of water =300g

temparature= 25°c

Change in temperature =(25-x)

Let us suppose 'x' is the final temperature

Cold body(Ice)

mass of cold body = 100g

Temparature of ice given as 0°c

Latent heat of ice = 80 cal/g

Water at 25°c converts into water at x°c

Then heat absorbed = mc(25-x)

Ice at 0°c converts into water at 0°c and water at 0°c coverts into water at x°c

Heat gained = mL+mc (x-0)

According to principle of calorimetry

Heat absorbed = Heat gained

mc(25-x)=mL+mc (x-0)

300 (1)(25-x)=100×80+100 (0.47)(x-0)

7500-300x=8000+47x

-500=347x

x= -1.44

If the final temperature 'x' is negative we need to assume the final temperature as 0°c

-Student of

*PhysicsWallahAlakhpandey*