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8 days ago

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.

Physics

1 answer:

Ostrovityanka [942]8 days ago

4 0

Answer:

b. 9.5°C

Explanation:

$m_i$ = Ice mass = 50 g

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

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

$m_w$ = Water mass = 200 g

$c_w$ = Water's specific heat = 4186 J/kg⋅°C

$m_{Al}$ = Aluminum mass = 80 g

$c_{Al}$ = Aluminum's specific heat = 900 J/kg⋅°C

The equation governing heat exchange in the system is presented 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 calculates to 9.50022°C

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11 days ago

Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.

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$Q=m C_p \Delta T$
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4 days ago

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