Question

One gram of ice at zero degree celsius is mixed with one gram of water at hundred degree celsius the resulting temperature will be

Answer 1

Explanation:

But the whole material come to an equilibrium state so the ice is converted into water at 100c. thus attains equilibrium. is, it takes 80 calories just to change 1 gram of ice at 0°C to 1 gram of water at 0°C. water at 20° with 1 g.

Answer 2

The resulting temperature will be $15.9^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$     ....(1)

where,

q = heat absorbed or released

$m_1$ = mass of ice= 1 g

$m_2$ = mass of water = 1 g

$T_{final}$ = final temperature = ?

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

$T_2$ = temperature of water = $100^oC$

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

$c_2$ = specific heat of water= $4.18J/g^0C$

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

$1\times 2.1\times (T_{final}-0)=-[1\times 4.18\times (T_{final}-100)]$

$T_{final}=15.9^0C$

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