Plan and Solve What unknown are you trying to calculate? Amount of energy absorbed, Q What formula contains the given quantity and the unknown? Q = mcAT = mc(T-T₁) Substitute the known values and solve. Q=(25.0 g)(0.48 cal/g-°C)(0.0°C - (-10.0°C)) = 120 cal Look Back and Check Is your answer reasonable? Yes, the sign shows that heat was absorbed and the unit, c proper unit for energy. Math Practice Note that

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Chapter1: Units, Trigonometry. And Vectors
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**Plan and Solve**

**What unknown are you trying to calculate?**  
Amount of energy absorbed, Q

**What formula contains the given quantity and the unknown?**  
\[ Q = mc\Delta T = mc(T_f - T_i) \]

**Substitute the known values and solve.**  
\[ Q = (25.0 \, \text{g})(0.48 \, \text{cal/g°C})(0.0°C - (-10.0°C)) = 120 \, \text{cal} \]

**Look Back and Check**  
**Is your answer reasonable?**  
Yes, the sign shows that heat was absorbed and the unit, calories, is a proper unit for energy.

---

**Math Practice**

On a separate sheet of paper, solve the following problems. Note that quantity of heat energy required to change the phase of a substance is given by the following equation: \( Q = mL \), where \( Q \) is the heat required, \( m \) is the mass, and \( L \) is the heat of fusion or vaporization.

1. **How much energy is absorbed when 25.0 g of ice at 0.0°C melts to form water at 0.0°C?** The heat of fusion of ice is 80 cal/g.

2. **How much energy is absorbed when the water at 0.0°C in Question 1 is heated to 100.0°C?** The specific heat of water is 1.0 cal/g°C.

3. **How much energy is absorbed when the water at 100.0°C in Question 2 vaporizes to 100.0°C?** The heat of vaporization of water is 540 cal/g.

4. **The steam in Question 3 is used to melt ice. How much ice will be melted by the energy released by the condensation of the steam at 100.0°C?**

---

(Note: The handwritten note at the bottom says "#4 I please," indicating a request for help with question 4.)
Transcribed Image Text:**Plan and Solve** **What unknown are you trying to calculate?** Amount of energy absorbed, Q **What formula contains the given quantity and the unknown?** \[ Q = mc\Delta T = mc(T_f - T_i) \] **Substitute the known values and solve.** \[ Q = (25.0 \, \text{g})(0.48 \, \text{cal/g°C})(0.0°C - (-10.0°C)) = 120 \, \text{cal} \] **Look Back and Check** **Is your answer reasonable?** Yes, the sign shows that heat was absorbed and the unit, calories, is a proper unit for energy. --- **Math Practice** On a separate sheet of paper, solve the following problems. Note that quantity of heat energy required to change the phase of a substance is given by the following equation: \( Q = mL \), where \( Q \) is the heat required, \( m \) is the mass, and \( L \) is the heat of fusion or vaporization. 1. **How much energy is absorbed when 25.0 g of ice at 0.0°C melts to form water at 0.0°C?** The heat of fusion of ice is 80 cal/g. 2. **How much energy is absorbed when the water at 0.0°C in Question 1 is heated to 100.0°C?** The specific heat of water is 1.0 cal/g°C. 3. **How much energy is absorbed when the water at 100.0°C in Question 2 vaporizes to 100.0°C?** The heat of vaporization of water is 540 cal/g. 4. **The steam in Question 3 is used to melt ice. How much ice will be melted by the energy released by the condensation of the steam at 100.0°C?** --- (Note: The handwritten note at the bottom says "#4 I please," indicating a request for help with question 4.)
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