When a 50.0 g sample of a metal at 100.0 ° C is added tO₆0.0 g of water at 25.10 ° C, the final temperature of both the water and metal is 32.81 ° 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 the metal equals the heat gained by the water. Using the formula $ q = mc\Delta T $, we can set up the equation: $ (50.0 \, \text{g})(c)(100.0 - 32.81) = (60.0 \, \text{g})(4.184 \, \text{J/g} \cdot \text{K})(32.81 - 25.10) $. Solving for $ c $ gives approximately 0.576 J/g × K.

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