Question

1. calculate the specific heat capacity of a piece of wood if 1625.9 g of the wood absorbs 76,500 joules of heat, and its temperature changes from 32°c to 75°c.

1. calculate the mass of a piece of gold that absorbs 12,500 joules and its temperature changes from 44°c to 97°c if the specific heat of gold is 0.129 j/g °c.

1. calculate the change in temperature if 1200 joules of heat is absorbed by 455 g of silver. the specific heat of silver is 0.237 j/g °c.

Explanation:

Problem 4

Step1: Recall the heat formula

The formula for heat absorbed or released is \( Q = mc\Delta T \), where \( Q \) is heat, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. We need to solve for \( c \), so rearrange the formula to \( c=\frac{Q}{m\Delta T} \).

Step2: Calculate \( \Delta T \)

\( \Delta T = T_{final}-T_{initial}=75^{\circ}C - 32^{\circ}C = 43^{\circ}C \).

Step3: Substitute values into the formula

Given \( Q = 76500 \, \text{J} \), \( m = 1625.9 \, \text{g} \), \( \Delta T = 43^{\circ}C \). Plug into \( c=\frac{Q}{m\Delta T} \): \( c=\frac{76500}{1625.9\times43} \).
First, calculate the denominator: \( 1625.9\times43 = 1625.9\times40+1625.9\times3 = 65036 + 4877.7 = 69913.7 \).
Then, \( c=\frac{76500}{69913.7}\approx1.094 \, \text{J/g}^{\circ}\text{C} \).

Step1: Recall the heat formula and rearrange for \( m \)

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

Step2: Calculate \( \Delta T \)

\( \Delta T = 97^{\circ}C - 44^{\circ}C = 53^{\circ}C \).

Step3: Substitute values into the formula

Given \( Q = 12500 \, \text{J} \), \( c = 0.129 \, \text{J/g}^{\circ}\text{C} \), \( \Delta T = 53^{\circ}C \). Plug into \( m=\frac{Q}{c\Delta T} \): \( m=\frac{12500}{0.129\times53} \).
First, calculate the denominator: \( 0.129\times53 = 6.837 \).
Then, \( m=\frac{12500}{6.837}\approx1828 \, \text{g} \).

Step1: Recall the heat formula and rearrange for \( \Delta T \)

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

Step2: Substitute values into the formula

Given \( Q = 1200 \, \text{J} \), \( m = 455 \, \text{g} \), \( c = 0.237 \, \text{J/g}^{\circ}\text{C} \). Plug into \( \Delta T=\frac{Q}{mc} \): \( \Delta T=\frac{1200}{455\times0.237} \).
First, calculate the denominator: \( 455\times0.237 = 107.835 \).
Then, \( \Delta T=\frac{1200}{107.835}\approx11.13 \, ^{\circ}\text{C} \).

Answer:

The specific heat capacity of the wood is approximately \( 1.09 \, \text{J/g}^{\circ}\text{C} \) (rounded to three significant figures).

Problem 5