12// 9.00 g of aluminum at 200C and 22.0 g of copper are dropped into 46.0 cm3 o

Question

12// 9.00 g of aluminum at 200C and 22.0 g of copper are dropped into 46.0 cm3 of ethyl alcohol at 15C. The temperature quickly comes to 27.0 C.
Part A
What was the initial temperature of the copper?

tem 16
A 12 m ×14m house is built on a 10 cm -thick concrete slab.
Part A
What is the heat-loss rate through the slab if the ground temperature is 9 while the interior of the house is 25 ?
Express your answer using two significant figures.

13/ The burner on an electric stove has a power output of 2.0kW. A 620 g stainless steel tea kettle is filled with 20C water and placed on the already hot burner.

tem 14
A 150 L(40gal) electric hot-water tank has a 5.0 kW heater.
Part A
How many minutes will it take to raise the water temperature from 62.0 F to 132 F ?

Part A

If it takes 2.30 min for the water to reach a boil, what volume of water, in cm3, was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.

Explanation / Answer

16.solution

The thermal conductivity of concrete varies widely. The value for dense concrete is 1.40 W/m*K.

H = Q/t = k*A*T/x
A = 168 m²
T = 25 - 9 = 16 K
x = 0.10 m

H = 1.4*168*16/0.10 = 37.632 kW

Low density concrete has a cp = 0.38 W/m*K, and in that case H = 6.32 kW

13.solution

The heat energy, which is produced by electric stove, is equal to the heat energy, which is required to increase the temperature of the tea kettle and water from 20C to 100C.
1st determine the amount of heat, which is required to increase the temperature of the tea kettle and water.
Q = mass in kg * Specific heat * T
Specific heat of iron = 450 J/kg
Q = 0.620 * 450 * 80
Specific heat of water = 4180 J/kg
Q = m * 4180 * 80

Total heat energy required = 0.620 * 450 * 80 + m * 4180 * 80

2nd determine the amount of heat energy, which the electric stove supplied.
Power = Energy ÷ time
Energy = Power in watts * time in seconds
Power = 2000
Time = 2.30 * 60
Energy supplied = 2000 * 2.30 * 60

Total heat energy required = Energy supplied
0.620 * 450 * 80 + m * 4180 * 80 = 2000 * 2.30 * 60
Solve for mass of the water.
1 kg of water = 1 liter of water

m = (276000 - 22320) / 334400
m = 0.758 litres

14.solution

Convert temperatures to Celsius and find the difference:
55.55 - 16.66 = 38.89 °C
150 liters with density = 1 kg/liter then mass is 150 kg
m = 150 kg
Cp = 4,186 kJ/kg K
Total heat is: Q = m * Cp * DeltaT = 150 * 4.186 * 38.89 = 24419 kJ

Heater is 5 kW or 5 kJ/sec

In 1 second we get 5 kJ
in x seconds we get 24419 kJ

x = 4883.8 sec = 81.39 min = 1 hour 21 minutes

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