Question

50 g ice at 0c is dropped in a calorimeter of negligible heat capacity containing 50 g water at 100c, then final temperature (tm) of the mixture is (neglect the heat loss)

Answer 1

Answer:

final temperature is 0c

Answer 2

The final temperature of the mixture is $66^oC$

Explanation:

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

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= 50 g

$m_2$ = mass of water = 50 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of ice= 0°C

$T_2$ = initial temperature of water = 100°C

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

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

Putting values in equation 1, we get:

$50\times 2.1\times (T_{final}-0)=-[50\times 4.186\times (T_{final}-100)]$

$T_{final}=66^oC$

Hence, the final temperature of the mixture is $66^oC$

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