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
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