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Delicious77 [7]
3 years ago
14

14. Heat energy equal to 25,000 J is applied to a 1200 g brick whose specific heat is 2.45 J/gºC.

Physics
1 answer:
choli [55]3 years ago
4 0

Answer:

∆T = 51 ºC

T2 = 76 ºC

Explanation:

(a)

∆Q = 25,000 J

m = 1200 g

c = 2.45 J/gºC

∆T = ?

∆Q = mc∆T

∆T = ∆Q/mc

∆T = 25000/(1200×2.45)

∆T = 51 ºC

(b) If T1 = 25 C

∆T = T2 - T1

T2 = ∆T + T1

T2 = 51+25

T2 = 76 ºC

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Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use
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Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

              Total Energy before fall = Total Energy after fall

                                E_{i}  = E_{f}

Here, total energy is the sum of kinetic energy and potential energy

K.E_{i} + P.E_{i} = K.E_{f} + P.E_{f}       (a)

When banana is at height h, it has

                 K.E_{i} = 0    and    P.E_{i} = mgh          

and when it reaches the river, it has

       K.E_{f}  = 1/2mv^{2}    and   P.E_{f}  = 0

Putting the values in equation (a)

                              0 + mgh = 1/2mv^{2} + 0

                                      mgh = 1/2mv^{2}

<em>cutting 'm' from both sides</em>

<em>                                           </em>gh = 1/2v^{2}

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5 0
3 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.
Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

P_1 = Initial pressure = 55.1 mmHg

P_2 = Final pressure = 1 atm = 760 mmHg

T_2 = Boiling point

T_1 = Initial temperature = 35°C

\Delta H_{vap} = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K

The normal boiling point of the substance is 389.78681 K

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The answer is 9.8, did this last year in AP Science

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