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Ch.5 - Thermochemistry

Chapter 5, Problem 106d

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

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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

A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs approximately 1.8 kg. The specific heat of the brick is 0.85 J/g•K. How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.7⨉103 gal of water?

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Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (a) Determine the amount of heat, in J, lost by the copper block.

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Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (b) Determine the amount of heat gained by the water. The specific heat of water is 4.184 J/1gK.

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We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethane: 2 CH4(g) → C2H6(g) + H2(g) Calculate the ΔH° for this reaction using the following thermochemical data: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH° = -890.3 kJ 2 H2(g) + O2(g) → 2 H2O(l) H° = -571.6 kJ 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ΔH° = -3120.8 kJ

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