Question

Imagine that you have two cups, one containing 1540 g of water
and the other one containing some amount of an unknown liquid. The
specific heat of water is 4.19 J/g·°C, and the specific heat of the
unknown liquid is 7.03 J/g·°C. You use two identical immersion
heaters to heat the water and the unknown liquid simultaneously.
Assume that all the heat from the heater is used to heat the
content of the cups.

(a) You started heating the water and the unknown liquid at the
same time. If the temperature of the water was increased from 23.7
°C to 40.8 °C while the temperature of the unknown liquid is
increased from 23.7 to 31.5 °C, what was the mass of the unknown
liquid in the cup?

________-grams

(b) If you used 102 W heaters to heat the liquids in part (a), how
long did you heat the liquids?

________-minutes

------------------------------------------------------------------

You and your friend went to kitchen to boil water. The specific
heat of water is 4.19 J/g °C, and the latent heat of vaporization
of water is 2260 J/g.

(a) You poured 179 gram of water to a container and put the
container on a 212 W electric heater. After turning on the heater,
you went away for 22 minutes to do something else. The container
was not covered so that the vaporized water was removed from the
container. When you came back to the kitchen 22 minutes later to
check the water, the container had 83.0 gram of water left in it
boiling. What was the temperature of your water when when you
turned on the heater?

________-°C

(b) Your friend wanted her own boiling water. She poured 426 gram
of water into another container and put her own electric heater at
the same time as you did. The initial temperature of her water was
the same as yours. When you came back to the kitchen to check the
water in 22 minutes, there was 301.2 gram of water left in her
container. What was the power of your friend's electric
heater?

__________-W

------------------------------------------------

Consider a mass lifting heat engine like the one you
experimented with in the lab. The pressure, volume, and temperature
data for the heat engine are shown in the table below:

.

State | Pressure | Volume | Temperature [°C] |
---|---|---|---|

a | p_{a} |
v_{a} |
2.0 |

b | p_{b} |
v_{b} |
2.0 |

c | p_{c} |
v_{c} |
40.0 |

d | p_{d} |
v_{d} |
40.0 |

a´=a | p_{a} |
v_{a} |
2.0 |

In the above table, p_{a} = p_{d} and p_{b}
= p_{c}. Also v_{d} - v_{a} = v_{c}
- v_{b}.

The diameter of the piston is is measured to be 14.15 mm.

a) If the work done by the engine in one cycle is 6.0 mJ, and the
pressure difference between the state **b** and
**d** is 5.0 kPa, what would be the volume change as
the engine moves from the state **d** to
**a**?

a) 1.2 m^{3} b) 1.2 ml c) 1.2x10^{-3} ml

b) Then, how high was the object lifted?

a) 0.0076 m b) 0.76 m c) 3.1 m

Answer #1

I will do only 4 parts,

a) Heat supplied is same

( 1540)(4.19) ( 40.8- 23.7) = m ( 7.03) ( 31.5- 23.7)

mass of unknown liquids= 2012.245 gm apprx

b) power = heat/time

time= ( 110339.46 J/ 102) =18.029 mon

c) let the initial tem of water = T

heat suplied = P x T= 212 ( 22x 60 ) = 279840 J

279840 = 179 x 4.19 ( 100 -T) + 2260 ( 179- 83)

179 ( 4.19 ) ( 100-T) = 62880

100-T = 62880/ 750.01

100-T = 83.83

T = 16.16 degree C

d) heat = 426 ( 4.19 ) ( 100- 16.16 ) + 124.8 ( 2260) = 431697. 3696 J

power = ( 431697. 3696 J)/ (22x 60 )= 327 W

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