When a 40.0 g sample of a metal at 25.00 ° C is added tO₆5.0 g of water at 100.00 ° C, the final temperature of both the water and metal is 93.27 ° C. The specific heat of water is 4.184 J/g × K. What is the specific heat of the metal?

Question

Quizplus · Accepted Answer

The heat lost by water equals the heat gained by the metal. Using the formula $ q = mc\Delta T $, solve for the specific heat of the metal: $ (65.0 \, \text{g})(4.184 \, \text{J/g} \cdot \text{K})(100.00 - 93.27) = (40.0 \, \text{g})(c)(93.27 - 25.00) $. Solving for $ c $ gives approximately 0.670 J/g × K.

When a 400 G Sample of a Metal at 25