The cooling system in an automobile holds 10.0 L of ethylene glycol antifreeze.How much energy is absorbed from the engine when the temperature of the ethylene glycol goes from 20.0$\degree$C to 100.0$\degree$C? The density and specific heat capacity of ethylene glycol are 1.11 g/mL and 2.42 J/(g . $\degree$C), respectively.

Question

Quizplus · Accepted Answer

To find the energy absorbed, use the formula $q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. First, convert the volume of ethylene glycol to mass using its density: $10.0 \, L = 10000 \, mL$, and since the density is $1.11 \, g/mL$, the mass $m = 10000 \, mL \times 1.11 \, g/mL = 11100 \, g$. The temperature change $\Delta T = 100.0^\circ C - 20.0^\circ C = 80.0^\circ C$. Thus, $q = 11100 \, g \times 2.42 \, J/(g\cdot^\circ C) \times 80.0^\circ C = 2150160 \, J = 2150 \, kJ$.

The Cooling System in an Automobile Holds 10 °\degree° C to 1000

The Cooling System in an Automobile Holds 10 $\degree$ C to 1000