Question

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).

Answer 1

The volume of the olive oil is 0.468 L.

Step-by-step 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
`\rho=860\ kg/m^3`

Specific heat of lead
`c_(l)= 128\ J/kg^(\circ)C`

Specific heat of olive oil
`c_(o)= 1970\ J/kg^(\circ)C`

We need to calculate the mass of the olive oil

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

`m_(1)c_(1)(T_(1)-T_(3))=m_(2)c_(2)(T_(3)-T_(2))`

`2.07*128(92-41)=m_(2)*1970*(41-24)`

`13512.96=m_(2)*33490`

`m_(2)=(13512.96)/(33490)`

`m_(2)=0.403\ kg`

We need to calculate the volume of olive oil

Using formula of density of olive oil

`\rho = (m)/(V)`

`V=(m)/(\rho)`

Put the value into the formula

`V=(0.403)/(860)`

`V=0.000468\ m^3`

`V=0.468*10^(-3)\ m^3`

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