Question

calculate the energy required to heat 0.70 kg of water from 35.1 °c to 48.1 °c. assume the specific heat capacity of water under these conditions is 4.18 j·g⁻¹·k⁻¹. round your answer to 2 significant digits.

Explanation:

Step1: Convert mass to grams

The mass of water is \( 0.70 \, \text{kg} \). Since \( 1 \, \text{kg} = 1000 \, \text{g} \), we convert the mass:
\( m = 0.70 \, \text{kg} \times 1000 \, \text{g/kg} = 700 \, \text{g} \)

Step2: Calculate temperature change

The initial temperature \( T_1 = 35.1^\circ \text{C} \) and the final temperature \( T_2 = 48.1^\circ \text{C} \). The temperature change \( \Delta T \) is:
\( \Delta T = T_2 - T_1 = 48.1^\circ \text{C} - 35.1^\circ \text{C} = 13.0^\circ \text{C} \) (Note: A change in Celsius is the same as a change in Kelvin, so \( \Delta T \) in K is also 13.0 K)

Step3: Use the heat formula \( q = mc\Delta T \)

The specific heat capacity \( c = 4.18 \, \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1} \), mass \( m = 700 \, \text{g} \), and \( \Delta T = 13.0 \, \text{K} \). Plugging in the values:
\( q = 700 \, \text{g} \times 4.18 \, \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1} \times 13.0 \, \text{K} \)
First, calculate \( 700 \times 4.18 = 2926 \)
Then, \( 2926 \times 13.0 = 38038 \, \text{J} \)

Step4: Round to 2 significant digits

The value \( 38038 \, \text{J} \) rounded to 2 significant digits is \( 3.8 \times 10^4 \, \text{J} \) (or \( 38000 \, \text{J} \) when considering significant figures, but in scientific notation it's clearer as \( 3.8 \times 10^4 \, \text{J} \))

Answer:

\( 3.8 \times 10^4 \, \text{J} \) (or \( 38000 \, \text{J} \))