Question

1. a 40 g sample of water absorbs 500 joules of energy. how much did the water temperature change? the specific heat of water (liquid) is 4.18 j/(g*°c).

Explanation:

Step1: Write the heat - absorption formula

The formula for heat absorption is \(Q = mc\Delta T\), where \(Q\) is the heat absorbed, \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. We need to solve for \(\Delta T\), so we can rewrite the formula as \(\Delta T=\frac{Q}{mc}\).

Step2: Substitute the given values

Given that \(Q = 500\space J\), \(m=40\space g\), and \(c = 4.18\space J/(g\cdot^{\circ}C)\).
Substitute these values into the formula: \(\Delta T=\frac{500}{40\times4.18}\).
First, calculate the denominator \(40\times4.18 = 167.2\).
Then, \(\Delta T=\frac{500}{167.2}\approx 2.99^{\circ}C\).

Answer:

The temperature change of the water is approximately \(3.0^{\circ}C\).