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Rainbow [258]
3 years ago
10

The value of specific heat for copper is 390 J/kg⋅°C, for aluminun is 900 J/kg⋅°C, and for water is 4186 J/kg⋅°C. What will be t

he equilibrium temperature when a 265 g block of copper at 235°C is placed in a 135g aluminum calorimeter cup containing 865 g of water at 14.0°C? Express your answer using three significant figures.
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

the equilibrium temperature Te = 19.9°C

Explanation:

Given;

specific heat for copper Cc is 390 J/kg⋅°C

for aluminun Ca is 900 J/kg⋅°C,

for water Cw is 4186 J/kg⋅°C

Mass of copper Mc= 265 g = 0.265kg

Temperature of copper Tc = 235°C

Mass of aluminium Ma = 135g = 0.135 kg

Temperature of aluminium Ta = 14.0°C

Mass of water Mw= 865 g = 0.865kg

Temperature of water Tw = 14.0°C

The equilibrium temperature can be derived by;

Te = (MaCaTa + McCcTc + MwCwTw)/(MaCa + McCc+ MwTw)

Substituting the values;

Te = ( 0.135×900×14 + 0.265×390×235 + 0.865×4186×14)/(0.135×900 + 0.265×390 + 0.865×4186)

Te = 19.939°C

Te = 19.9°C

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A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 \Omega. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300\Omega . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

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    The  open circuit voltage is  V =  40.0V

     The internal resistance is R = 2 \Omega

     The emf of each battery is e =  6.00 V

      The internal resistance of the battery is  r = 0.300V

      The  charging current is  I = 4.00 \ A

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        The total resistance in the circuit is

                              R_T =  R + 2r +R_z

Substituting values

                             R_T = 2.6 +R_z

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                         E = V -2e

                 =>   E =  40 - (2 * 6)

                        E =  28 V

Now according to ohm's law

            I = \frac{E}{R_T}

Substituting values

           4 = \frac{28}{R_z + 2.6}        

Making R_z the subject of the formula

So    R_z =  \frac{28 - 10.4}{4}

           R_z =  4.4 \Omega

The  increase rate of   internal energy at the supply is mathematically represented as

        Z_1  = I^2 R

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         Z_2 = I^2 r

Substituting values

         Z_2 = 4^2 * 2 * 0.3

         Z_2 = 9.6 \ W

The  increase rate of  internal energy at the added  series resistance  is mathematically represented as

        Z_3 = I^2 R_z

Substituting values

       Z_3 = 4^2 * 4.4

      Z_3 = 70.4 W

Generally the increase rate of the chemically energy in the battery is  mathematically represented as

         C = 2 * e * I

Substituting values

       C =  2 * 6  * 4

      C =  48 W

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