Math Problem Statement

A car engine's cooling system con- tains 20.0 L of water (1 L of water has a mass of 1 kg).

A) What is the change in the temperature of the water if 836.0 kJ of thermal energy is addled?

b) Suppose that it is winter, and the car's cooling system is filled with methanol. The density of methanol is 0.80 g/cm", What would be the increase in temperature of the methanol if it absorbed 836.0 kJ of thermal energy?

Solution

Let's break this problem down into two parts.

Part A: Change in temperature of water

We need to find the change in temperature (ΔT\Delta T) of the water when 836.0 kJ of thermal energy is added.

We can use the formula for heat transfer:

Q=mcΔTQ = mc\Delta T

Where:

  • QQ is the thermal energy added (836.0 kJ = 836×103836 \times 10^3 J),
  • mm is the mass of the water,
  • cc is the specific heat capacity of water (cwater=4186J/kgCc_{\text{water}} = 4186 \, \text{J/kg}^\circ C),
  • ΔT\Delta T is the change in temperature (which we are solving for).

Given:

  • Volume of water = 20.0 L,
  • 1 L of water has a mass of 1 kg, so mass mwater=20.0kgm_{\text{water}} = 20.0 \, \text{kg}.

Substituting these values into the equation:

836×103=20.0×4186×ΔT836 \times 10^3 = 20.0 \times 4186 \times \Delta T

Solving for ΔT\Delta T:

ΔT=836×10320.0×4186\Delta T = \frac{836 \times 10^3}{20.0 \times 4186}

Let me calculate that.

Part B: Change in temperature of methanol

For the methanol case, we are given the density of methanol and need to calculate the mass of methanol first.

  • The density of methanol is 0.80 g/cm³, which is equivalent to 0.80 kg/L.
  • Volume of methanol = 20.0 L (since the same cooling system is being filled with methanol).

The mass of methanol is:

mmethanol=density×volume=0.80kg/L×20.0L=16.0kgm_{\text{methanol}} = \text{density} \times \text{volume} = 0.80 \, \text{kg/L} \times 20.0 \, \text{L} = 16.0 \, \text{kg}

Now, using the heat transfer equation again, but this time with methanol's specific heat capacity:

  • Q=836×103Q = 836 \times 10^3 J,
  • mmethanol=16.0kgm_{\text{methanol}} = 16.0 \, \text{kg},
  • Specific heat capacity of methanol cmethanol=2.51kJ/kgC=2510J/kgCc_{\text{methanol}} = 2.51 \, \text{kJ/kg}^\circ C = 2510 \, \text{J/kg}^\circ C.

Using the same formula for heat transfer:

836×103=16.0×2510×ΔT836 \times 10^3 = 16.0 \times 2510 \times \Delta T

Solving for ΔT\Delta T:

ΔT=836×10316.0×2510\Delta T = \frac{836 \times 10^3}{16.0 \times 2510}

Let me calculate both ΔT\Delta T values now.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can walk you through the calculation steps manually so you can compute the answer yourself.

Would you like more details on the steps? Here are some related questions to think about:

  1. What would happen if the water volume were increased to 30 L?
  2. How does the specific heat capacity of methanol compare to water in terms of thermal stability?
  3. How does the density of a substance affect the overall mass in thermal calculations?
  4. What role does thermal energy play in internal combustion engines beyond the cooling system?
  5. How would these results differ in an insulated vs. a non-insulated system?

Tip: When solving heat transfer problems, always ensure that units are consistent to avoid errors.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Heat Transfer
Thermodynamics
Algebra

Formulas

Q = mcΔT (Heat transfer equation)
Density = Mass/Volume

Theorems

Law of Conservation of Energy

Suitable Grade Level

Grades 10-12