Question

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

What is the final equilibrium temperature? (Specific heat for aluminum is 900 J/kg⋅°C, the specific heat of water is 4 186 J/kg⋅°C, and Lf = 3.33 × 105 J/kg.)
a. 17.9°Cb. 9.5°Cc. 12.1°Cd. 20.6°C

Answer 1

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

Answer 2

b. 9.5°C is the final equilibrium temperature

What is equilibrium in Physics?

Equilibrium, in physics, is the condition of a system when neither its state of motion nor its internal energy state tends to change with time.

According to the question the given values are:

Mass of ice ($m_{i}$) = 50 g ,

Initial temperature of water and Aluminum ($T_{i}$ ) = 30°C

Latent heat of fusion ( $L_{f}$)= $3.33*10^{5} J/kg$ °C

Mass of water( $m_{w}$) = 200 g

Specific heat of water ( $c_{w}$) = 4186 J/kg °C

Mass of Aluminum ( $m_{al}$) = 80 g

Specific heat of Aluminum ( $c_{al}$) = 900 J/kg  °C

The  heat exchange equation of the systems is:

$m_{i} (L_{f} + c_{w} T) + m_{w} c_{w} (T- T_{i} )+ m_{al} c_{al} =0$

By putting all values in the equation:

T = $\frac{10626}{1118.5} = 9.50022$

T = 9.50022 °C

Therefore,

The final equilibrium temperature is 9.50022 °C

Learn more about equilibrium here:

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