A 500.0 grams sample of aluminum is initially at 25.0C it absorbs 32.60 KJ of heat from its surroundings. What is it final temperature, in C?

72·45 °C

Explanation:

Given

Mass of sample of aluminum is 500·0 g

Initial temperature is 25·0 °C = 298 K

Heat absorbed from surroundings is 32·60 kJ

Heat capacity of aluminum is 0·9 J/g K

Formula for amount of heat transfer is

Q = m * s * ΔT

where

Q is the amount of heat transfer

m is the mass of the substance

s is the heat capacity of the substance

ΔT is the change in temperature

Let the final temperature be T K

∴ 32·60 * 1000 = 500 * 0·9 * (T - 298)

T - 298 = 72·45

∴ T = 370·45 K = (370·45 - 298) °C = 72·45 °C

∴ Final temperature is 72·45 °C