How much heat energy is needed to change 95 g of water at 25^circ C into steam at 100^circ C ? Use 4.2(J)/(gcdot ^circ )C for the specific heat of water. Use 2260J/g for the heat of vaporization of water. (1 point) 2.1times 10^5 3.0times 10^4 1.8times 10^4 2.4times 10^5

Here's how to calculate the total heat energy required:<br /><br />**1. Heating the water from 25°C to 100°C:**<br /><br />* Temperature change: $\Delta T = 100^{\circ}C - 25^{\circ}C = 75^{\circ}C$<br />* Mass of water: $m = 95g$<br />* Specific heat of water: $c = 4.2 \frac{J}{g \cdot ^{\circ}C}$<br />* Heat required (Q1): $Q1 = m \cdot c \cdot \Delta T = 95g \cdot 4.2 \frac{J}{g \cdot ^{\circ}C} \cdot 75^{\circ}C = 29925J$<br /><br />**2. Vaporizing the water at 100°C:**<br /><br />* Mass of water: $m = 95g$<br />* Heat of vaporization: $L_v = 2260 \frac{J}{g}$<br />* Heat required (Q2): $Q2 = m \cdot L_v = 95g \cdot 2260 \frac{J}{g} = 214700J$<br /><br />**3. Total heat required:**<br /><br />* Total heat (Q): $Q = Q1 + Q2 = 29925J + 214700J = 244625J$<br /><br />Since the answer choices are given in scientific notation, we can express this as $2.44625 \times 10^5 J$. Rounding to two significant figures, we get $2.4 \times 10^5 J$.<br /><br />Therefore, the correct answer is $\boxed{2.4\times 10^{5}}$.<br />