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You wish to cool a 2.07 kg block of lead initially at 92.0°C to a temperature of 41.0°C by placing it in a container of olive oil initially at 24.0°C. Determine the volume (in L) of the liquid needed in order to accomplish this task without boiling. The density and specific heat of olive oil are respectively 860 kg/m3 and 1,970 J/(kg · °C), and the specific heat of lead is 128 J/(kg · °C).

Respuesta :

Answer:

The volume of the olive oil is 0.468 L.

Explanation:

Given that,

Mass of lead m₁= 2.07 kg

Initial temperature of lead = 92.0°C

Initial temperature of oil = 41.0°C

Density of olive oil [tex]\rho=860\ kg/m^3[/tex]

Specific heat of lead [tex]c_{l}= 128\ J/kg^{\circ}C[/tex]

Specific heat of olive oil [tex]c_{o}= 1970\ J/kg^{\circ}C[/tex]

We need to calculate the mass of the olive oil

The heat lost by the lead equal to the heat gained by olive oil

[tex]m_{1}c_{1}(T_{1}-T_{3})=m_{2}c_{2}(T_{3}-T_{2})[/tex]

[tex]2.07\times128(92-41)=m_{2}\times1970\times(41-24)[/tex]

[tex]13512.96=m_{2}\times33490[/tex]

[tex]m_{2}=\dfrac{13512.96}{33490}[/tex]

[tex]m_{2}=0.403\ kg[/tex]

We need to calculate the volume of olive oil

Using formula of density of olive oil

[tex]\rho = \dfrac{m}{V}[/tex]

[tex]V=\dfrac{m}{\rho}[/tex]

Put the value into the formula

[tex]V=\dfrac{0.403}{860}[/tex]

[tex]V=0.000468\ m^3[/tex]

[tex]V=0.468\times10^{-3}\ m^3[/tex]

Hence, The volume of the olive oil is 0.468 L.

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