Question

If 20 g steam initially at 1000C is added to 60 g of ice initially at 00C, then find the final

equilibrium temperature of the mixture​

Answer 1

Answer:

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Explanation:

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Answer 2

The final  equilibrium temperature of the mixture​ is $24.0^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$heat_{absorbed}=heat_{released}$

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

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

where,

Q = heat absorbed or released

$m_1$ = mass of ice = 60 g

$m_2$ = mass of steam = 20 g

$T_{final}$  = final temperature = ?

$T_1$ = temperature of ice =$0^0C$

$T_2 = temperature of steam = 100^oC$

$c_1$ = specific heat of ice = $2.1J/g^0C$

$c_2$ = specific heat of steam= $1.99J/g^0C$

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

$T_{final}=24.0^0C$

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