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 to accomplish this task without boiling. The density and specific heat of kerosene are 820 kg/m³ and 2,010 J/(kg·°C) respectively, and the specific heat of tin is 218 J/(kg·°C).

To cool the tin block, approximately 0.4878 liters of kerosene are needed. This is determined by equating the heat lost by the tin to the heat gained by the kerosene and solving for the kerosene's mass and volume.

Explanation:

The subject question asks us to determine the volume of kerosene needed to cool a block of tin from 88.0°C to 49.0°C without boiling the kerosene. To solve this problem, we apply the principle of conservation of energy, which means the heat lost by the tin will equal the heat gained by the kerosene. The specific heat formula Q = mcΔT is used, where Q is the heat energy transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, we calculate the heat lost by the tin:

Qtin = mtin × ctin × (ΔTtin)

Qtin = 1.59 kg × 218 J/(kg°C) × (88.0°C - 49.0°C)

Qtin = 1.59 kg × 218 J/(kg°C) × 39°C

Qtin = 13745.82 J

Next, we find the mass of kerosene needed using the heat gained formula:

Qkerosene = mkerosene × ckerosene × (ΔTkerosene)

13745.82 J = mkerosene × 2010 J/(kg°C) × (49.0°C - 32.0°C)

mkerosene = 13745.82 J / (2010 J/(kg°C) × 17°C)

mkerosene = 0.40 kg

Finally, we convert the mass of kerosene to volume:

Vkerosene = mkerosene / ρkerosene

Vkerosene = 0.40 kg / 820 kg/m³

Vkerosene = 0.0004878 m³

Vkerosene = 0.4878 L

Therefore, approximately 0.4878 liters of kerosene are needed to cool the block of tin to the desired temperature.

thomasdecabooter thomasdecabooter · Answer 2 · 2024-09-06T00:53:06-04:00

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