Question

A 1.00×1 0 2 g aluminum block at 100.0°C is placed in 1.00×1 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?

CA=(-mBCB∆TB)/mA∆TA

Answer 1

The answer to your question is Ca = 0.904 J/g°C

Step-by-step explanation:

Data

Aluminum Water

mass 1 x10² g 1 x 10² g

Temperature 1 100°C 26°C

Temperature 2 10°C 10°C

Specific heat ? 4.182 J/g°C

Formula

Aluminum Water

-mCa(T2 - T1) = mCw(T2 - T1)

-Solve for Ca

Ca = [mCw(T2 - T1)] / -m(T2 - T1)

-Substitution

Ca = [1 x 10² x 4.182 x (26 - 10] / -1 x10²(26 - 100)

-Simplification

Ca = [418.2(16)] / -1 x 10²(-74)

Ca = 6691.2 / 7400

-Result

Ca = 0.904 J/g°C