Question

The specific heat of ethanol is 2.44 J/g ֯C. How many kJ of energy are required to heat 50.0 grams of ethanol from -20 ֯C to 68 ֯C? (heat equation)

Answer 1

To heat 50.0 grams of ethanol from -20°C to 68°C, 10.752 kJ of energy are required, calculated using the specific heat of ethanol and the temperature change.

Step-by-step explanation:

To find the energy required to heat 50.0 grams of ethanol from -20 °C to 68 °C, you would use the formula:
q = mcΔT, where:

• q is the heat energy in joules (J),
• m is the mass in grams (g),
• c is the specific heat in J/g°C, and
• ΔT is the change in temperature in °C (ΔT = final temperature - initial temperature).

Here, m = 50.0 g, c = 2.44 J/g°C, and ΔT = (68 - (-20)) °C = 88 °C. Plugging these values into the formula gives:

q = (50.0 g) × (2.44 J/g°C) × (88 °C) = 10,752 J

To convert joules to kilojoules (kJ), we divide by 1,000:

q = 10,752 J ÷ 1,000 = 10.752 kJ

Therefore, 10.752 kJ of energy is required to heat 50.0 grams of ethanol from -20 °C to 68 °C.

Answer 2

Heat energy required (Q) = 10.736 KJ

Step-by-step explanation:

Given:

Specific heat of ethanol (C) = 2.44 J/g °C

Mass of ethanol (M) = 50 gram

Initial temperature (T1) = -20°C

Final temperature (T1) = 68°C

Find:

Heat energy required (Q) = ?

Computation:

Change in temperature (ΔT) = 68°C - (-20°C)

Change in temperature (ΔT) = 88°C

Heat energy required (Q) = mC(ΔT)

Heat energy required (Q) = (50)(2.44)(88)

Heat energy required (Q) = 10,736 J

Heat energy required (Q) = 10.736 KJ