Question

If liquid mercury, Hg, has a specific heat of 0.14 J/g°C, how much heat would be required to heat 75 grams of Hg (ℓ) from 22°C to 28°C? Report your answer in joules to the nearest whole number.

a. 63 J
b. 82 J
c. 105 J
d. 119 J

Answer 1

To heat 75 grams of mercury from 22°C to 28°C, we use the formula Q = mcΔT, with a specific heat of 0.14 J/g°C for mercury, resulting in a required heat of 63 joules.

Step-by-step explanation:

To calculate the heat required to raise the temperature of a substance, we can use the formula:

Q = mcΔT

Where:

• Q is the heat in joules (J)
• m is the mass of the substance in grams (g)
• c is the specific heat capacity (J/g°C)
• ΔT is the change in temperature (°C)

In this case, we have the following values:

• m = 75 g
• c = 0.14 J/g°C for mercury
• ΔT = 28°C - 22°C = 6°C

Let's plug these values into the formula:

Q = (75 g)(0.14 J/g°C)(6°C)

Now, multiplying these together:

Q = 75 x 0.14 x 6 = 63 J

Therefore, the heat required to heat 75 grams of mercury from 22°C to 28°C is 63 J.