Question

Suppose you store a 15.5 g piece of aluminum (Cp of Al = 0.902 J/g⁰C) in the refrigerator at 2.3⁰C and then drop it into your coffee. The coffee temperature drops from 95.6⁰C to 45.6⁰C. How many Joules of energy did the aluminum block absorb?

Answer 1

605.4 J

Step-by-step explanation:

When a certain substance absorbs a certain amount of energy, its temperature increases according to the equation:

`Q=mC\Delta T`

where

Q is the heat absorbed

m is the mass of the substance

C is the specific heat capacity of the substance

`\Delta T` is the change in temperature

In this problem, we have:

m = 15.5 g is the mass of the piece of aluminium

C = 0.902 J/g⁰C is the specific heat of the aluminium

`\Delta T = 45.6-2.3=43.3^(\circ)C` is the change in temperature of the aluminium (in fact, at thermal equilibrium, the block of aluminium reaches the same final temperature as the coffee)

Therefore, the energy absorbed is

`Q=(15.5)(0.902)(43.3)=605.4 J`