Question

A 27.7 g sample of ethylene glycol, car radiator coolant,loses 688 J of heat. What is the initial temperature of the ethylene glycol if the finaltemperature is 32.5 ºC (specific heat, c, of ethylene glycol = 2.42 J/g·K?)

Answer 1

The initial temperature of the ethylene glycol is approximately 22.94 ºC.

Step-by-step explanation:

To find the initial temperature of the ethylene glycol, we can use the formula:

q = mc∆T

where:

• q is the heat lost (-688 J)
• m is the mass of the sample (27.7 g)
• c is the specific heat of ethylene glycol (2.42 J/g·K)
• ∆T is the change in temperature (final temperature - initial temperature)

Since the final temperature is given as 32.5 ºC, we can substitute the known values into the equation and solve for the initial temperature:

-688 J = (27.7 g)(2.42 J/g·K)(32.5 ºC - initial temperature)

Simplifying the equation, we have:

-688 J = 66.9346 g·K(32.5 ºC - initial temperature)

Now, we can isolate the initial temperature:

-688 J = 66.9346 g·K(32.5 ºC) - 66.9346 g·K(initial temperature)

-688 J = 2174.990 g·K - 66.9346 g·K(initial temperature)

Solving for initial temperature:

Initial temperature = (2174.990 g·K - 688 J) / -66.9346 g·K

Using a calculator, we find that the initial temperature of the ethylene glycol is approximately 22.94 ºC.