Question

A sample of water has a mass of 0.625 kg. The thermal energy of the water sample changes by 105,000 J as its temperature is raised from 25°C to 65°C. Find the specific heat of water to two significant digits.

Answer 1

To find the specific heat capacity of water, you can use the formula:

\[
q = m \cdot c \cdot \Delta T
\]

where:
- \( q \) is the heat energy transferred (in joules),
- \( m \) is the mass of the substance (in kilograms),
- \( c \) is the specific heat capacity (in J/kg·°C),
- \( \Delta T \) is the change in temperature (in °C).

Given:
- \( q = 105,000 \, \text{J} \)
- \( m = 0.625 \, \text{kg} \)
- Initial temperature \( T_1 = 25 \, \text{°C} \)
- Final temperature \( T_2 = 65 \, \text{°C} \)

First, we can calculate the change in temperature:

\[
\Delta T = T_2 - T_1 = 65 \, \text{°C} - 25 \, \text{°C} = 40 \, \text{°C}
\]

Now, we can rearrange the formula to solve for the specific heat capacity \( c \):

\[
c = \frac{q}{m \cdot \Delta T}
\]

Substituting the given values into the equation:

\[
c = \frac{105,000 \, \text{J}}{0.625 \, \text{kg} \cdot 40 \, \text{°C}}
\]

Now, calculate \( c \):

\[
c = \frac{105,000}{0.625 \cdot 40}
\]

Calculating the denominator:

\[
0.625 \cdot 40 = 25
\]

Now substituting back:

\[
c = \frac{105,000}{25} = 4,200 \, \text{J/kg·°C}
\]

To two significant digits, the specific heat of water is:

\[
\boxed{4200} \, \text{J/kg·°C}
\]