Thank you for visiting In the laboratory a student finds that it takes 103 Joules to increase the temperature of 12 6 grams of solid diamond from 22 4. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer:
The correct solution is "[tex]0.480 \ J/g^{\circ}C[/tex]".
Explanation:
Given:
q = 103 J
Mass,
m = 12.6 grams
Temperature,
[tex]T_1=22.4[/tex]
[tex]T_2=39.4[/tex]
[tex]\Delta T=T_2-T_1[/tex]
[tex]=39.4-22.4[/tex]
[tex]=17^{\circ}C[/tex]
Now,
⇒ [tex]C=\frac{q}{m\times \Delta T}[/tex]
⇒ [tex]=\frac{103}{12.6\times 17}[/tex]
⇒ [tex]=\frac{103}{214.2}[/tex]
⇒ [tex]=0.480 \ J/g^{\circ}C[/tex]
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