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babymother [125]
3 years ago
12

A steel casting weighing 2 kg has an initial temperature of 500°c; 25 kg of water initially at 25°c is contained in a perfectly

insulated steel tank weighing 5 kg. the casting is immersed in the water and the system is allowed to come to equilibrium. what is its final temperature? ignore the effects of expansion or contraction, and assume constant specific heats of 4.18 kj⋅kg−1⋅°c−1 for water and 0.50 kj⋅kg−1⋅°c−1 for steel.
Physics
2 answers:
gayaneshka [121]3 years ago
5 0

Answer:

26.6C

Explanation:

Using an energy balance:

Decrease in internal energy of casting must lead to an increase in internal energy of the tank and water, assuming that no heat flows out of the tank - perfectly insulated.

m_{casting}*C_{p,casting}*(T_{casting} - T_{final})= m_{water}*C_{p,water}*(T_{final} - T_{water}) + m_{tank}*C_{p,tank}*(T_{final} - T_{tank})

2*0.5*(500 - T_{final})= 70*4.18*(T_{final} - 25) + 5*0.5*(T_{final} - 25)\\\\(1+2.5+292.6)*T_{final} = 500+7315+62.5\\\\T_{final} = 26.6C

SCORPION-xisa [38]3 years ago
3 0
<span> Plan: Use Q = m · c · ΔT three times. Hot casting cools ΔT_hot = 500°C - Tf. Cold water and steel tank heat ΔT_cold = Tf - 25°C. Set Q from hot casting cooling = Q from cold tank heating.
here
m_cast · c_steel · ΔT_hot = (m_tank · c_steel + m_water · c_water) · ΔT_cold

m_cast · c_steel · (500°C - Tf) = (m_tank · c_steel + m_water · c_water) · (Tf - 25°C)

2.5 kg · 0.50 kJ/(kg K°) · (500°C - Tf) = (5 kg· 0.50 kJ/(kg K°) + 40 kg· 4.18 kJ/(kg K°)) · (Tf - 25°C)

Solve for Tf, remember that K° = C° (i.e. for ΔT's) </span>
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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

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Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

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See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


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