You wish to cool a 1.59 kg block of tin initially at 88.0°C to a temperature of 49.0°C by placing it in a container of kerosene initially at 32.0°C. Determine the volume (in L) of the liquid needed in order to accomplish this task without boiling. The density and specific heat of kerosene are respectively 820 kg/m3 and 2,010 J/(kg · °C), and the specific heat of tin is 218 J/(kg · °C).

Question

contestada

Respuesta :

thomasdecabooter thomasdecabooter · Answer 1 · 2019-12-21T04:26:41+01:00

Answer:

We need 0.482 L of kerosene

Explanation:

Step 1: Data given

Mass of the block tin = 1.59 kg

Initial temperature tin= 88.0 °C

Final temperature = 49.0 °C

Initial temperature of kerosene = 32.0 °C

Density of kerosene = 820 kg/m³

Specific heat of tin is 218 J/(kg · °C)

Step 2: Calculate mass of kerosene

Heat lost = heat gained

Qtin = -Qkerosene

Q = m*c*ΔT

m(tin) * c(tin) * ΔT(tin) = -m(kerosene) * c(kerosene) * ΔT(kerosene)

⇒ mass of tine = 1.59 kg

⇒ c(tin) = the specific heat of tin = 218 J/ kg*°C

⇒ ΔT(tin) = The change in temperature = T2 - T1 = 49.0 °C - 88.0 °C = -39.0 °C

⇒ mass of kerosene = TO BE DETERMINED

⇒ c(kerosene) = The specific heat of kerosene = 2010 J/kg*°C

⇒ ΔT = 49.0 - 32.0 = 17.0

1.59 kg * 218 J/kg*°C * -39.0 °C = - m(kerosene) * 2010 J/kg*°C *17.0 °C

-13518.18 = -m(kerosene) * 34170

m(kerosene) = 0.39562 kg

Step 3: Calculate volume of kerosene

Volume = mass / density

Volume = 0.39562 kg / 820 kg/m³

Volume = 4.82 * 10^-4 m³ = 0.482 L

We need 0.482 L of kerosene