To determine the mass of a water sample that absorbs 1450 J of energy while experiencing a temperature increase of 6.5°C, we can use the formula that relates energy absorbed (q), mass (m), specific heat capacity (c), and change in temperature (ΔT):
[tex]\[ q = mc\Delta T \][/tex]
Given:
- [tex]\( q \)[/tex] (energy absorbed) = 1450 J
- [tex]\( \Delta T \)[/tex] (change in temperature) = 6.5°C
- [tex]\( c \)[/tex] (specific heat capacity of water) = 4.18 \frac{J}{g \cdot °C}
We need to solve for the mass [tex]\( m \)[/tex]. Rearranging the formula to solve for [tex]\( m \)[/tex]:
[tex]\[ m = \frac{q}{c\Delta T} \][/tex]
Substitute the given values into the equation:
[tex]\[ m = \frac{1450}{4.18 \times 6.5} \][/tex]
Upon performing the division and multiplication:
- Calculate the product of the specific heat capacity and the change in temperature:
[tex]\[ 4.18 \times 6.5 = 27.17 \][/tex]
- Now, divide the energy absorbed by this product:
[tex]\[ m = \frac{1450}{27.17} \approx 53.368 \][/tex]
Thus, the mass of the water sample is approximately 53.368 grams.
