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:

Question 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 change in temperature. We need to solve for \( c \), so rearrange the formula to \( c=\frac{Q}{m\Delta T} \).

Step2: Calculate \( \Delta T \)

\( \Delta T = T_2 - T_1 = 75^\circ C - 32^\circ C = 43^\circ C \).

Step3: Substitute values into the formula

Given \( Q = 76500 \, J \), \( m = 1625.9 \, g \), \( \Delta T = 43^\circ C \). So \( c=\frac{76500}{1625.9\times43} \). First calculate the denominator: \( 1625.9\times43 \approx 1625.9\times40 + 1625.9\times3 = 65036 + 4877.7 = 69913.7 \). Then \( c=\frac{76500}{69913.7} \approx 1.094 \, J/g^\circ C \).

Step1: Recall the heat formula

The formula is \( Q = mc\Delta T \), we need to solve for \( m \), so rearrange to \( m=\frac{Q}{c\Delta T} \).

Step2: Calculate \( \Delta T \)

\( \Delta T = T_2 - T_1 = 97^\circ C - 44^\circ C = 53^\circ C \).

Step3: Substitute values into the formula

Given \( Q = 12500 \, J \), \( c = 0.129 \, J/g^\circ C \), \( \Delta T = 53^\circ C \). So \( m=\frac{12500}{0.129\times53} \). First calculate the denominator: \( 0.129\times53 = 6.837 \). Then \( m=\frac{12500}{6.837} \approx 1828.3 \, g \).

Step1: Recall the heat formula

The formula is \( Q = mc\Delta T \), we need to solve for \( \Delta T \), so rearrange to \( \Delta T=\frac{Q}{mc} \).

Step2: Substitute values into the formula

Given \( Q = 1200 \, J \), \( m = 455 \, g \), \( c = 0.237 \, J/g^\circ C \). So \( \Delta T=\frac{1200}{455\times0.237} \). First calculate the denominator: \( 455\times0.237 \approx 455\times0.2 + 455\times0.037 = 91 + 16.835 = 107.835 \). Then \( \Delta T=\frac{1200}{107.835} \approx 11.13^\circ C \).

Answer:

Approximately \( 1.09 \, J/g^\circ C \) (or more precisely \( \approx 1.094 \, J/g^\circ C \))

Question 5