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:

\[ \boxed{4200} , \text{J/kg·°C} \]