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aev [14]
2 years ago
9

Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not cha

nge. However, if a significant amount of heat flows from the object to the thermometer, the temperature will change. A thermometer has a mass of 33.0 g, a specific heat capacity of c = 804 J/(kg C°), and a temperature of 15.4 °C. It is immersed in 149 g of water, and the final temperature of the water and thermometer is 56.0 °C. What was the temperature of the water in degrees Celsius before the insertion of the thermometer?
Physics
1 answer:
nordsb [41]2 years ago
5 0

Answer : The temperature of water is, 57.7^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of thermometer = 804J/kg.^oC=0.804J/g^oC

c_2 = specific heat of water = 4.18J/g.^oC

m_1 = mass of thermometer = 33.0 g

m_2 = mass of water = 149 g

T_f = final temperature = 56.0^oC

T_1 = initial temperature of thermometer = 15.4^oC

T_2 = initial temperature of water = ?

Now put all the given values in the above formula, we get:

33.0g\times 0.804J/g^oC\times (56.0-15.4)^oC=-149g\times 4.18J/g.^oC\times (56.0-T_2)^oC

T_2=57.7^oC

Therefore, the temperature of water is, 57.7^oC

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Part 1 :
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Answer:

1. 218.55 N

2. 30.96^{o}

3. 2.1 m/s^{2}

Explanation:

Part 1;

Net force F=mg sin \theta where m is mass, g is gravitational force and \theta is the angle of inclination

F= 46*9.8*sin 29^{o}= 218.55N

Frictional force, F_{r} is given by

F_{r} = \mu_{s}mg cos \theta where \mu_{s} is the coefficient of static friction

F_{r} = 0.6*46*9.8 cos 29

F_{r}=236.57N

Since F_{r}>F, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force F_{s} = \mu_{s} mg cos \theta

mg sin \theta =\mu_{s} mg cos \theta

\mu_{s}= \frac {sin \theta}{cos \theta}

\mu_{s}= tan \theta

The maximum angle of inclination \theta = tan^{-1} \mu_{s}

\theta = tan^{-1} (0.6)

\theta= 30.96^{o}  

        

Part 3:

Net force on the object is given by

ma = mg sin 38 - \mu_{k} mg cos 38 where \mu_{k} is the coefficient of kinetic friction

 a = g ( sin 38 - \mu_{k} cos 38 )

                 = 9.8 ( sin 38 - (0.51) cos 38 )

                = 2.1m/s^{2}

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3 years ago
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