find the initial temperature of water, (e), (k), (l) a hot metal is placed inside the calorimeter containing water. Inside the calorimeter, the hotmetal releases heat, which in turn is absorbed by the water and the calorimeter. The heatreleased by the metal is equal to the heat absorbed by the calorimeter and the water basedon the equation shown:Heat released (Q) = Heat absorbed (Q)Heat released by metal = Heat absorbed by the calorinmeter + Heat absorbed by waterQmetalQa + Qqrater(mretal Cmatal ATmeta)= (m.a Ca ATa) + (mate Cwatur ATrater)= mass in gramsC = specific heat of a substance(theoretical)?AT = change in temperature (Tf-Ti)Where:mTable 4.1. Specific Heat of Common Metals.Specific Heat – (C)(Cal/g°C)0.220Metal(J/g®C)0.910AluminumBrassCopperIron (Fe)LeadMagnesiumTin0.3800.3900.0900.0920.4600.1080.1300.0311.0500.3820.2430.0541.000Water4.184 Results, and calculations: Complete the table below, using Fe as the metal20g105g85gWt. of CalorimeterWt. of Calorimeter + waterWeight of WaterInitial temperature of water, (t)v°C(e) Initial temperature of aluminum calorimeter, (t)eal(f) Initial temperature of metal (t)metal(g) Weight of metal sample, mmetal(h) Final temperature of water, (t)w(i) Final temperature of calorimeter, (t)calG) Final temperature of metal(k) Heat absorbed by the metal, gcal(1) Heat absorbed by the water, gw°C100 °C26 g30.5 °C30.5 °C30.5 °CJouleJoule(m) Experimental specific heat of metal sample0.492 J/gC(n) Theoretical specific heat of metal sample0.460 J/gC(0) Percent Error7.0 %

We have to predict the initial temperature of calorimeter , heat absorbed by Calorimeter and water.…

Answered: find the initial temperature of water,…

Science

Chemistry

find the initial temperature of water, (e), (k), (l)

World of Chemistry

7th Edition

ISBN:9780618562763

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter10: Energy

Section: Chapter Questions

Problem 65A

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Question

find the initial temperature of water, (e), (k), (l)

a hot metal is placed inside the calorimeter containing water. Inside the calorimeter, the hot
metal releases heat, which in turn is absorbed by the water and the calorimeter. The heat
released by the metal is equal to the heat absorbed by the calorimeter and the water based
on the equation shown:
Heat released (Q) = Heat absorbed (Q)
Heat released by metal = Heat absorbed by the calorinmeter + Heat absorbed by water
Qmetal
Qa + Qqrater
(mretal Cmatal ATmeta)
= (m.a Ca ATa) + (mate Cwatur ATrater)
= mass in grams
C = specific heat of a substance(theoretical)?
AT = change in temperature (Tf-Ti)
Where:
m
Table 4.1. Specific Heat of Common Metals.
Specific Heat – (C)
(Cal/g°C)
0.220
Metal
(J/g®C)
0.910
Aluminum
Brass
Copper
Iron (Fe)
Lead
Magnesium
Tin
0.380
0.390
0.090
0.092
0.460
0.108
0.130
0.031
1.050
0.382
0.243
0.054
1.000
Water
4.184

Results, and calculations: Complete the table below, using Fe as the metal
20g
105g
85g
Wt. of Calorimeter
Wt. of Calorimeter + water
Weight of Water
Initial temperature of water, (t)v
°C
(e) Initial temperature of aluminum calorimeter, (t)eal
(f) Initial temperature of metal (t)metal
(g) Weight of metal sample, mmetal
(h) Final temperature of water, (t)w
(i) Final temperature of calorimeter, (t)cal
G) Final temperature of metal
(k) Heat absorbed by the metal, gcal
(1) Heat absorbed by the water, gw
°C
100 °C
26 g
30.5 °C
30.5 °C
30.5 °C
Joule
Joule
(m) Experimental specific heat of metal sample
0.492 J/gC
(n) Theoretical specific heat of metal sample
0.460 J/gC
(0) Percent Error
7.0 %

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