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Ch.9 - Thermochemistry: Chemical Energy

Chapter 9, Problem 153c

(c) Assume that a chunk of potassium weighing 7.55 g is dropped into 400.0 g of water at 25.0 °C. What is the final temperature of the water if all the heat released is used to warm the water?

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Hi everyone. So we have a one kg block of nickel metal And it's headed to 98°C. That was dropped in a coffee cup calorie meter containing 350 g of water? At 25.8°C. The temperature inside the cup became 41.3°C. We were asked if the heat loss from the metal was all absorbed by the water. What would be the final temperature of the system? All that the amount of heat transferred is even to the mass of the solution. Talk to specific heat. The solution times the temperature change. They're gonna be looking at the heat of nettle equal to the mass of nickel. That's a specific heat, nickel. How's the temperature change for a mask? Love one kg And in one kg 1000 g. So this is 1000 g. And we're looking for the heat of the nickel and the specific key capacity of nickel 0.44 course jewels are grams times degrees Celsius. And for the temperature change 41. Grace Celsius minus 98 Very Celsius. We get negative 56.7 Grace Celsius. I'll be plugging advice to get the heat of nickel was 1000 g From 0. doors programs. Times degrees Celsius. I was gonna get 56.7 Celsius. The heat of the medical 25, 174 0.8 jaws. So now we need to find the final temperature of the solution. Have the heat of water to the mass of water specific heat capacity of the water. How's the temperature change for the heat of the water? 25174. I ate jaws, the mask, 350 grams for the specific heat capacity of the water. 4.184. Those programs, times degrees Celsius for the temperature change and the final temperature -25.8°C. Then I can plug in advice to find the final temperature. I have 25,000 174 0. doors. It was 350 grams £4.184. Our grams times degrees Celsius. The final temperature -25.8°C When I was 25,074. Eight jaws. Just 1464 point forward yours three degrees Celsius. That was the final temperature -25.8. MS Celsius Created by both sides by 1,464. What for those part degree Celsius at 17.19 cassius, It's the final temperature -25.8. Very Celsius, Hope you at 25.8 resale system. Both sides. For the final temperature. They get 43 C. Thanks for watching my video and I hope it was helpful
Related Practice
Textbook Question

Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are

(1) CH3CO2H(l) + Na2CO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)

(2) CH3CO2H(l) + NaHCO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)

(b) How many kilograms of each substance is needed to neutralize a 1.000-gallon spill of pure acetic acid (density = 1.049 g/mL)?

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Textbook Question

(a) Write a balanced equation for the reaction of potassium metal with water.

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Textbook Question

(b) Use the data in Appendix B to calculate ΔH° for the reaction of potassium metal with water.

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Textbook Question

(d) What is the molarity of the KOH solution prepared in part (c), and how many milliliters of 0.554 M H2SO4 are required to neutralize it?

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Textbook Question
Hydrazine, a component of rocket fuel, undergoes combus- tion to yield N2 and H2O: N2H41l2 + O21g2 S N21g2 + 2 H2O1l2 (b) Use the following information to set up a Hess's law cycle, and then calculate ΔH° for the combustion reac- tion. You will need to use fractional coefficients for some equations. 2 NH31g2 + 3 N2O1g2 S 4 N21g2 + 3 H2O1l2 ΔH° = - 1011.2 kJ N2O1g2 + 3 H21g2 S N2H41l2 + H2O1l2 ΔH° = - 317.2 kJ 4 NH31g2 + O21g2 S 2 N2H41l2 + 2 H2O1l2 ΔH° = - 286.0 kJ H2O1l2 ΔH°f = - 285.8 kJ>mol
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Textbook Question
Reaction of gaseous fluorine with compound X yields a sin- gle product Y, whose mass percent composition is 61.7% F and 38.3% Cl. (c) Calculate ΔH° for the synthesis of Y using the following information: 2 CIF1g2 + O21g2 S Cl2O1g2 + OF21g2 ΔH° = + 205.4 kJ 2 CIF31l2 + 2 O21g2 S Cl2O1g2 + 3 OF21g2 ΔH° = + 532.8 kJ OF21g2 ΔH°f = + 24.5 kJ>mol
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