Question

1. a 15.75 - g piece of iron absorbs 1086 j of heat energy, and its temperature changes from 25°c to 175°c. calculate the specific heat capacity of iron.
2. what mass of water will change its temperature by 3.0°c when 525 j of heat is added to it?
3. 250 kj of energy are removed from a 500 g sample of water at 60°c. will the sample of water completely freeze? explain.

Explanation:

6.

Step1: Identify the formula

The formula for heat energy is $Q = mc\Delta T$, where $Q$ is heat energy, $m$ is mass, $c$ is specific - heat capacity, and $\Delta T$ is the change in temperature. We need to solve for $c$, so $c=\frac{Q}{m\Delta T}$.

Step2: Calculate $\Delta T$

$\Delta T=T_2 - T_1=175^{\circ}C - 25^{\circ}C = 150^{\circ}C$.

Step3: Substitute values into the formula

Given $Q = 1086\ J$, $m = 15.75\ g$, and $\Delta T=150^{\circ}C$. Then $c=\frac{1086\ J}{15.75\ g\times150^{\circ}C}$.
$c=\frac{1086}{2362.5}\ J/(g\cdot^{\circ}C)\approx0.46\ J/(g\cdot^{\circ}C)$.

Step1: Use the heat - energy formula

From $Q = mc\Delta T$, we can solve for $m$. The specific - heat capacity of water $c = 4.18\ J/(g\cdot^{\circ}C)$, $Q = 525\ J$, and $\Delta T = 3.0^{\circ}C$.

Step2: Rearrange the formula for $m$

$m=\frac{Q}{c\Delta T}$.

Step3: Substitute values

$m=\frac{525\ J}{4.18\ J/(g\cdot^{\circ}C)\times3.0^{\circ}C}=\frac{525}{12.54}\ g\approx42\ g$.

Step1: Calculate the energy required to cool water to $0^{\circ}C$

The specific - heat capacity of water $c = 4.18\ J/(g\cdot^{\circ}C)$, $m = 500\ g$, $\Delta T=60^{\circ}C$. Using $Q_1=mc\Delta T$, we have $Q_1 = 500\ g\times4.18\ J/(g\cdot^{\circ}C)\times60^{\circ}C=500\times4.18\times60\ J = 125400\ J=125.4\ kJ$.

Step2: Calculate the energy required to freeze water at $0^{\circ}C$

The heat of fusion of water $L_f = 334\ J/g$. For $m = 500\ g$, the energy required to freeze is $Q_2=mL_f=500\ g\times334\ J/g = 167000\ J = 167\ kJ$.

Step3: Calculate the total energy required to freeze the water

$Q_{total}=Q_1 + Q_2=125.4\ kJ+167\ kJ = 292.4\ kJ$.

Step4: Compare with the given energy

Given $Q_{removed}=250\ kJ$. Since $250\ kJ<292.4\ kJ$, the water will not completely freeze.

Answer:

The specific - heat capacity of iron is approximately $0.46\ J/(g\cdot^{\circ}C)$.

7.