Class Management Help Homework #10 Begin Date: 3/12/2018 12:00:00 AM-Due Date: 4
ID: 2036786 • Letter: C
Question
Class Management Help Homework #10 Begin Date: 3/12/2018 12:00:00 AM-Due Date: 4/8/2018 11:59:00 PM End Date: 4/15/2018 12:00:00 AM (3%) Problem 33: Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. Substances Specific heat (c) Solids kcal/kg C Aluminum Concrete (Granite) Lead Glass Ice (average J/kg C 900 840 128 840 2090 0.215 0.20 0.0305 0.20 0.50 Liquids Water 4186 1.000 Gases Steam (100°C) 1520 (2020)0.363(0.482) Otheexpertta.com × How much heat transfer. in kilocalories, is needed to heat and melt 0.475 kg of lead, starting from 24.5°C? The melting point of lead is 327°C and its latent heat of fusion is 5.85 kcal/kg Grade Summary Deductions Potential 14% 8600 cos) ?| tan() acos) sin Submissions Attempts remaining: 3 (5% per attempt) cotanasin0 atanacotan sih0 coshtanh cotanh) detailed view 5% 590 Degrees Radians NO CE I DELI CLEAR Submit Feedback Feedback: 1 for a 2% deduction You may have forgotten to include the process of melting lead into give up Hints: 1 for a 200 deduction. Hints remaining: 0 The temperature of the lead is raised from its initial temperature to its melting point and then the lead is melted. Both processes require heat. your calculation. Submission History Answer Feedback Totals Hints -The temperature of the lead is raised from its initial temperature to its melting point and then the lead is melted. Both 5% 290 090 790 10.23 processes requireExplanation / Answer
mass = m = 0.475 kg
initial temperature = t1 = 24.5 degrees
final temperature = t2 = 327 ^0C
L = 5.85 kcal/ kg
total heat energy = m c dT + m L
Q = 0.475* 0.0305* (327- 24.5) + 0.475 * 5.85
Q = 7.16 Kcal