Question

what is the mass of a water sample which absorbs 1450 j of energy when its temperature is increased by 6.5 °c? (c_{h₂o} = 4.18 \frac{j}{g·°c}) m = ? g

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is change in temperature. We need to solve for \( m \), so rearrange the formula: \( m=\frac{q}{c\Delta T} \).

Step2: Identify known values

We know \( q = 1450\ J \), \( c = 4.18\ \frac{J}{g\cdot^\circ C} \), and \( \Delta T = 6.5\ ^\circ C \).

Step3: Substitute values into the formula

\( m=\frac{1450\ J}{4.18\ \frac{J}{g\cdot^\circ C}\times6.5\ ^\circ C} \). First calculate the denominator: \( 4.18\times6.5 = 27.17 \). Then \( m=\frac{1450}{27.17}\approx53.37 \) g.

Answer:

\( \approx53.4 \) g (or more precisely, the value calculated as above)