Question

100 gm ice at -60 degree celcius is mixed with 20 gm of water at 20 degree celcius .Calculate the temperature of equillibrium mixture and mixture content

Answer 1

heat lost + heat gained = 0  

500 x 4.18 ( T - 80) + 200 x 4.18 ( T - 25)=0  

divide by 4.18  

500 T - 40000 + 200 T - 5000 = 0  

700 T = 45000  

T = 64.3 °C

Answer 2

The final temperature of the mixture -37.26°C

Explanation:

When ice is mixed in water, the amount of heat released by ice will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

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

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of ice = 100 g

$m_2$ = mass of water = 20 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of ice = -60°C

$T_2$ = initial temperature of water = 20°C

$c_1$ = specific heat of ice = 2.108 J/g°C

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$100\times 2.108\times (T_{final}-(-60))=-[20\times 4.186\times (T_{final}-20)]$

$T_{final}=-37.26^oC$

Learn more about specific heat:

https://brainly.in/question/15018962

https://brainly.com/question/14721973

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