Question

1. water has a higher specific heat capacity (4186 j/kg.°c) than iron (450 j/kg.°c). imagine you have equal masses of water and iron, both starting at thermal equilibrium. if you supply 126 j of heat to each, what will happen to their temperatures?

a. they are no longer at thermal equilibrium; the water is at a higher temperature.
b. they are no longer at thermal equilibrium; the iron is at a higher temperature.
c. they remain at thermal equilibrium.
d. impossible to say without knowing their exact masses.

1. what is the efficiency of an ideal carnot engine operating between a reservoir in which ice and water coexist, and a reservoir in which water and steam coexist? the pressure is constant at 1.0 atm for both reservoirs.

a. 1.0%
b. 100%
c. 27%
d. 0.27%

1. a heat pump extracts 460 j of heat from outside and then exhaust 712 j into a room. how much work is required to achieve this?

a. 1172 j
b. 620 j
c. 252 j
d. impossible to determine without the pumps coefficient of performance.

Explanation:

Question 4

Step1: Use the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. We can rewrite it for \(\Delta T\) as \(\Delta T=\frac{Q}{mc}\). Given \(m_{water}=m_{iron} = m\) and \(Q_{water}=Q_{iron}=Q = 126\space J\).

Step2: Compare the temperature changes

For water, \(\Delta T_{water}=\frac{Q}{m c_{water}}\), and for iron, \(\Delta T_{iron}=\frac{Q}{m c_{iron}}\). Since \(c_{water}(4186\space J/kg\cdot^{\circ}C)>c_{iron}(450\space J/kg\cdot^{\circ}C)\), then \(\Delta T_{water}<\Delta T_{iron}\) (because \(Q\) and \(m\) are the same). If they started at the same temperature (thermal equilibrium initially), after heating, \(T_{iron}=T_{initial}+\Delta T_{iron}\) and \(T_{water}=T_{initial}+\Delta T_{water}\), so \(T_{iron}>T_{water}\).

Step1: Recall the temperatures of phase - change points

At \(1\space atm\), the temperature of the ice - water co - existence (cold reservoir) \(T_{c}=273\space K\) and the temperature of water - steam co - existence (hot reservoir) \(T_{h}=373\space K\).

Step2: Use the Carnot efficiency formula

The Carnot efficiency formula is \(\eta = 1-\frac{T_{c}}{T_{h}}\). Substitute \(T_{c} = 273\space K\) and \(T_{h}=373\space K\) into the formula: \(\eta=1-\frac{273}{373}\approx1 - 0.732=0.268\approx 27\%\)

Step1: Use the first law of thermodynamics for heat pumps

For a heat pump, the first law of thermodynamics is \(Q_{h}=Q_{c}+W\), where \(Q_{h}\) is the heat exhausted into the hot reservoir (room), \(Q_{c}\) is the heat extracted from the cold reservoir (outside), and \(W\) is the work done.

Step2: Solve for \(W\)

Rearrange the formula to \(W = Q_{h}-Q_{c}\). Given \(Q_{c}=460\space J\) and \(Q_{h}=712\space J\), then \(W=712 - 460=252\space J\)

Answer:

B. They are no longer at thermal equilibrium; the iron is at a higher temperature.

Question 5