QuestionDecember 20, 2025

6. Suppose 88 grams of water are cooled from 22^circ C to 16^circ C How much heat is given off? 7. Suppose 16 grams of water are cooled from 13^circ C to 1^circ C How much heat is given off? 8. Suppose 10 calories of heat are added to 10 grams of water. The temperature of the water will increase by how much? 9. Suppose 100 calories of heat are added to 10 grams of water. The temperature of the water will increase by how much? 10. Suppose 50 calories of heat are added to 10 grams of water at 10^circ C . What will the final temperature be?

6. Suppose 88 grams of water are cooled from 22^circ C to 16^circ C How much heat is given off? 7. Suppose 16 grams of water are cooled from 13^circ C to 1^circ C How much heat is given off? 8. Suppose 10 calories of heat are added to 10 grams of water. The temperature of the water will increase by how much? 9. Suppose 100 calories of heat are added to 10 grams of water. The temperature of the water will increase by how much? 10. Suppose 50 calories of heat are added to 10 grams of water at 10^circ C . What will the final temperature be?
6. Suppose 88 grams of water are cooled from 22^circ C to 16^circ C How much heat is given off? 7. Suppose 16 grams of water are cooled from 13^circ C to 1^circ C How much heat is given off? 8. Suppose 10 calories of heat are added to 10 grams of water. The temperature of the water will increase by how much? 9. Suppose 100 calories of heat are added to 10 grams of water. The temperature of the water will increase by how much? 10. Suppose 50 calories of heat are added to 10 grams of water at 10^circ C . What will the final temperature be?

Solution
4.3(346 votes)

Answer

6. -528 cal ### 7. -192 cal ### 8. 1^\circ C ### 9. 10^\circ C ### 10. 15^\circ C Explanation 1. Use the heat formula for temperature change The formula is q = mc\Delta T, where q is heat (cal), m is mass (g), c is specific heat (1\, \text{cal/g}^\circ\text{C} for water), and \Delta T is temperature change. --- Problem 6 2. Calculate \Delta T \Delta T = 16^\circ C - 22^\circ C = -6^\circ C 3. Substitute values q = 88 \times 1 \times (-6) = -528 cal --- Problem 7 4. Calculate \Delta T \Delta T = 1^\circ C - 13^\circ C = -12^\circ C 5. Substitute values q = 16 \times 1 \times (-12) = -192 cal --- Problem 8 6. Rearrange formula to solve for \Delta T \Delta T = \frac{q}{mc} 7. Substitute values \Delta T = \frac{10}{10 \times 1} = 1^\circ C --- Problem 9 8. Rearrange formula to solve for \Delta T \Delta T = \frac{100}{10 \times 1} = 10^\circ C --- Problem 10 9. Rearrange formula to solve for \Delta T \Delta T = \frac{50}{10 \times 1} = 5^\circ C 10. Add \Delta T to initial temperature Final temperature = 10^\circ C + 5^\circ C = 15^\circ C

Explanation

1. Use the heat formula for temperature change<br /> The formula is $q = mc\Delta T$, where $q$ is heat (cal), $m$ is mass (g), $c$ is specific heat ($1\, \text{cal/g}^\circ\text{C}$ for water), and $\Delta T$ is temperature change.<br /><br />---<br /><br /> Problem 6<br /><br />2. Calculate $\Delta T$<br /> $\Delta T = 16^\circ C - 22^\circ C = -6^\circ C$<br /><br />3. Substitute values<br /> $q = 88 \times 1 \times (-6) = -528$ cal<br /><br />---<br /><br /> Problem 7<br /><br />4. Calculate $\Delta T$<br /> $\Delta T = 1^\circ C - 13^\circ C = -12^\circ C$<br /><br />5. Substitute values<br /> $q = 16 \times 1 \times (-12) = -192$ cal<br /><br />---<br /><br /> Problem 8<br /><br />6. Rearrange formula to solve for $\Delta T$<br /> $\Delta T = \frac{q}{mc}$<br /><br />7. Substitute values<br /> $\Delta T = \frac{10}{10 \times 1} = 1^\circ C$<br /><br />---<br /><br /> Problem 9<br /><br />8. Rearrange formula to solve for $\Delta T$<br /> $\Delta T = \frac{100}{10 \times 1} = 10^\circ C$<br /><br />---<br /><br /> Problem 10<br /><br />9. Rearrange formula to solve for $\Delta T$<br /> $\Delta T = \frac{50}{10 \times 1} = 5^\circ C$<br /><br />10. Add $\Delta T$ to initial temperature<br /> Final temperature $= 10^\circ C + 5^\circ C = 15^\circ C$
Click to rate:

Similar Questions