Question

A 15.0 g block of aluminum at an initial temperature of 27.5°C absorbs 0.678 kJ of heat. What is the final temperature of the block? The specific heat of Al is 0.902 J/g°C.

a) 50.1°C
b) 38.6°C
c) 54.5°C
d) 68.6°C
e) 77.6°C

Answer 1

The final temperature of the 15.0 g aluminum block after absorbing 0.678 kJ of heat is approximately 77.4°C. Considering round-off errors, the closest provided answer choice is 77.6°C.

Step-by-step explanation:

To determine the final temperature of a 15.0 g block of aluminum that absorbs 0.678 kJ of heat, we use the formula for heat transfer Q = m*c*ΔT, where Q is heat energy, m is mass, c is the specific heat, and ΔT is the change in temperature. Given that the specific heat (c) of aluminum is 0.902 J/g°C, the original temperature is 27.5°C, and the heat absorbed is 0.678 kJ (which is 678 J), we can solve for the change in temperature (ΔT).

First, convert the heat energy from kJ to J (1 kJ = 1000 J), so Q = 678 J. Then, rearrange the formula to solve for ΔT: ΔT = Q / (m*c).

Substituting the values gives us ΔT = 678 J / (15.0 g * 0.902 J/g°C) ≈ 49.89°C. Adding this to the initial temperature of 27.5°C gives us the final temperature, which is 27.5°C + 49.89°C ≈ 77.4°C. However, since this is not one of the options provided, and considering possible round-off errors, the closest option is e) 77.6°C.