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description
show all you work, including units, on separate paper. follow the \problem solving method\.
this final exam is cumulative and covers material from the entire course.
instructions
from the list of choices, select the one best answer.
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question 61
1.8 points save answer
it takes 3,680 cal to raise the temperature of a 500 g sample from 20°c to 100°c. what is the specific heat of the sample?
0.057 cal/g°c
0.092 cal/g°c
0.124 cal/g°c
0.024 cal/g°c

Explanation:

Step1: Recall the heat formula

The formula for heat \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is temperature change.

Step2: Calculate temperature change

\( \Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80^\circ\text{C} \)

Step3: Rearrange formula for \( c \)

From \( Q = mc\Delta T \), solve for \( c \): \( c=\frac{Q}{m\Delta T} \)

Step4: Substitute values

\( Q = 3680 \, \text{cal} \), \( m = 500 \, \text{g} \), \( \Delta T = 80^\circ\text{C} \). So \( c=\frac{3680}{500\times80} \)

Step5: Compute the value

\( 500\times80 = 40000 \), then \( c=\frac{3680}{40000}=0.092 \, \text{cal/g}^\circ\text{C} \)

Answer:

0.092 cal/g°C (corresponding to the option "0.092 cal/g°C")