EXPERIMENT A An aluminum calorimeter with water is used to determine specific he

Question

EXPERIMENT A

An aluminum calorimeter with water is used to determine specific heat of US pennies. Pennies are heated to BP water, and placed in calorimeter

CALORIMETER

Aluminum cup: 500 g

Water: 200 mL

Initial temperature: 19 C

PENNIES

One roll of 50 pennies: 2.5 g each

Initial temperature: 100 C

FINAL SYSTEM: CALORIMETER+ PENNIES

Final temperature: 21.9 C

A) Specific heat of US penny is ____ J/kgC

EXPERIMENT B

A quantity of steam at 100 C is injected into a container with a 1 cf (0.03m3) block of ice at 0 C and produces water at 20 C. The amount of steam needed is ___kg

Explanation / Answer

This system we assume that there is no loss of heat to the environment

Qabsorbed = -Q hot

Q= m ce (Tf-T1)

Q hot= mp ce (Tf – To )           To= 100°C                Tf= 21.9°C      mp= 50 2.5 = 125 g

Qabsorbed = (mal Ce Al + m H2O ceH2O) ( Tf- T1)           T1=19°C

. cc (aluminium) = 0.215 cal/g °c

. ce (water) = 1 cal /g °C

.ce(pennies) = -(malCe Al+ m H2OceH2O) ( Tf- T1) / (mp (Tf – T0 ) )

. ce (pennies) =- (500 0.215 + 200 1) (21.9-19) /(125 (21.9-100))

. ce = 207.5 2.9/(125 78.1) = 601.75/9762.5

. ce = 0.0616 cal/ g °C

.B   We calculate the amount of heat (Q1) required to melt the ice and the amount of heat to water up (Q2) to 20°C

Qt= Q1 +Q2

Ice density

. d = m/ V              V= 0.03 m3    d (ice T= 0°C) = 0.92 g/cm3 = 0.92 103 Kg/m3

. m = d V   m = 0.92 103 0.03     m= 27.6 Kg

L = 3.33 105 J/Kg latent heat

Q1 = mL         Q = 27.6 3.33 105 = 9.19 106 J

Q2 = mH2O ce H2o (Tf-To)                   Tf=20°C        To= 0°C

. d = m/V               d( water a 0°C) = 0.9998 g/cm3 = 999.8 Kg/m3

. mH2O = d V         mH2O = 999.8 0.03 = 29.99 Kg

Q2 = 29.99 4186 ( 20 – 0)        Q2= 2.51 106 J

Qt= 9.19 106 + 2.51 106               Qt= 11.7 106 J

This is the heat that must be supplied to heat the ice and take it to liquid water at 20° C calculate the amount of steam you need to supply this energy

Qsteam = m Ce (Tf-To)                  Tf= 20°C      To= 100°C                  Ce= 0.48 cal/g °C = 2010 J/Kg °C

Qsteam= - Qt

.   m = Qt/ ( Ce (Tf-To))                  m= -11.7 106 /( 2010 (20-100)     m=11.7 106/ 1.608 105

.    m =   72.76 Kg of steam

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