Assignments: Soc-1523-99 University of Oklahoma X sapling earning com Vibisc ms/

Question

Assignments: Soc-1523-99 University of Oklahoma X sapling earning com Vibisc ms/mod/ibis/view.php?id 3204 pling gleaming com Dynamic Periodic Canvas Aa Sapling Learning Jump to... Sapling Learning I University of Oklahoma CHEM 1415. Spring17-CLIFFORD Activities and Due Dates HW.1 My Assignment 20/2017 11:55 PM &31/50 1/18/2017 o4:25 PM Gradebook Attempts Score Print Calculator Periodic Table Question 8 of 14 Map In another step of this lab, you record the m Calorimetry experiment of water (mwater) in a coffee cup as mmater 30.00 g What is the initial temperature of the water, Twater. in the coffee cup as shown to the left? Number 26 In the next step of the lab (as shown in the animation to the left) you place the hot piece of etal into the cool water Play the animation and record the final temperature of the water, water 2. Reset Continue Number 29 12 With the information above and the specific heat of water, 4.184 JMg "C), you calculate the specific heat, c, of the metal. Number Previous Check Answer Next +I Exit Logout Hell Resources OAssignment Information Available From: /16/2017 06:00 PM 1/20/2017 11:55 PM Due Date: Points Possible Grade Category Grad escription: 5 attempt, 0% penalty, Policies: Solutions After Due Date You can check your answers. You can view solutions after e due date. You have five attempts per question. There is no penalty for incorrect answers O eTextbook O Help With This Topic oweb Help & Videos OTechnical Support and Bug Reports

Explanation / Answer

Here the first step of question is missing but you can calculate this question as below procedure,

The energy given off by the metal cooling = energy gained by the water.

The formula for any sensible heat change is energy = mass * specific heat * temperature change

mass of metal * specific heat * temperature change of metal = mass of water * specific heat * temperature change of water

You have all the above except for the specific heat of the metal, substitute and solve for the specific heat of the metal.

Here's my understanding: You dropped a 45-g piece of metal at 82 C into 130 g of H2O at 26 C and the final temperature was 29 C.

Heat gained by H2O = (mass H2O)(SH H2O)(Tf - Ti) = (130.00 g)(4.184 J/g C)(29 C - 26 C) = 2107 J

Heat lost by metal = 2107 J = (mass metal)(SH metal)(Ti - Tf)
2107 = (45.00 g)(SH metal)(82 C - 30 C)
2107 = 3690 SH metal - 1350
3457 = 3690 SH metal
SH metal = 3457 / 3690 = 0.9368 J/g C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.