A lab assistant drops a 400.0-g piece of metal at $100.0 ^ { \circ } \mathrm { C }$ into a 100.0-g aluminum cup containing 500.0 g of water at $15.0 ^ { \circ } \mathrm { C }$ In a few minutes, she measures the final temperature of the system to be $40.0 ^ { \circ } \mathrm { C }$ What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is $900.0 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and that of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$

Question

A lab assistant drops a  400.0-g piece of metal at $100.0 ^ { \circ } \mathrm { C }$ into a  100.0-g aluminum cup containing  500.0 g  of water at $15.0 ^ { \circ } \mathrm { C }$ In a few minutes, she measures the final temperature of the system to be $40.0 ^ { \circ } \mathrm { C }$ What is the specific heat of the  400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is $900.0 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and that of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$

Quizplus · Accepted Answer

The heat lost by the metal equals the heat gained by the water and the aluminum cup. Using $q = mc\Delta T$, where $q$ is heat, $m$ is mass, $c$ is specific heat, and $\Delta T$ is the change in temperature, we can set up the equation: $$q_{\text{metal}} = q_{\text{water}} + q_{\text{aluminum}}$$ Substituting the given values and solving for the specific heat of the metal ($c_{\text{metal}}$), we find that $c_{\text{metal}}$ is approximately $2270 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$.

A Lab Assistant Drops a 400 100.0∘C100.0 ^ { \circ } \mathrm { C }100.0∘C

A Lab Assistant Drops a 400 $100.0 ^ { \circ } \mathrm { C }$