Question

calculating heat worksheet
$q = mc\delta t$, where $q$ = heat energy, $m$ = mass, and $\delta t$ = change in temp
remember, $\delta t = (t_{final} - t_{initial})$. show all work and proper units.

1. isolate each variable in the equation $q = mc\delta t$ algebraically.

1. a 21.65g piece of iron absorbs 1045.15 joules of heat energy, and its

temperature changes from 25°c to 125°c. calculate the specific heat
capacity of iron.

1. how many joules of heat are needed to raise the temperature of

14.7g of aluminum from 18°c to 58°c, if the specific heat of aluminum
is 0.90 j/g°c?

Explanation:

Problem 1: Isolate each variable in \( Q = mc\Delta T \)

Step 1: Isolate \( m \)

To isolate \( m \), divide both sides of \( Q = mc\Delta T \) by \( c\Delta T \).
\( m=\frac{Q}{c\Delta T} \)

Step 2: Isolate \( c \)

Divide both sides by \( m\Delta T \) to isolate \( c \).
\( c=\frac{Q}{m\Delta T} \)

Step 3: Isolate \( \Delta T \)

Divide both sides by \( mc \) to isolate \( \Delta T \).
\( \Delta T=\frac{Q}{mc} \)

Step 4: Isolate \( Q \) (already isolated, but confirm)

The original equation is \( Q = mc\Delta T \), so \( Q \) is already isolated.

Step 1: Identify known values

\( Q = 1045.15 \, \text{J} \), \( m = 21.65 \, \text{g} \), \( T_{\text{initial}} = 25^\circ\text{C} \), \( T_{\text{final}} = 125^\circ\text{C} \).

Step 2: Calculate \( \Delta T \)

\( \Delta T = T_{\text{final}} - T_{\text{initial}} = 125 - 25 = 100^\circ\text{C} \).

Step 3: Use \( c=\frac{Q}{m\Delta T} \)

Substitute values: \( c=\frac{1045.15}{21.65 \times 100} \).

Step 4: Compute \( c \)

\( 21.65 \times 100 = 2165 \); \( \frac{1045.15}{2165} \approx 0.483 \, \text{J/g}^\circ\text{C} \).

Step 1: Identify known values

\( m = 14.7 \, \text{g} \), \( c = 0.90 \, \text{J/g}^\circ\text{C} \), \( T_{\text{initial}} = 18^\circ\text{C} \), \( T_{\text{final}} = 58^\circ\text{C} \).

Step 2: Calculate \( \Delta T \)

\( \Delta T = 58 - 18 = 40^\circ\text{C} \).

Step 3: Use \( Q = mc\Delta T \)

Substitute values: \( Q = 14.7 \times 0.90 \times 40 \).

Step 4: Compute \( Q \)

\( 14.7 \times 0.90 = 13.23 \); \( 13.23 \times 40 = 529.2 \, \text{J} \).

Answer:

• \( m=\boldsymbol{\frac{Q}{c\Delta T}} \)
• \( c=\boldsymbol{\frac{Q}{m\Delta T}} \)
• \( \Delta T=\boldsymbol{\frac{Q}{mc}} \)
• \( Q=\boldsymbol{mc\Delta T} \)

Problem 2: Calculate specific heat capacity of iron