PHYSICS SET UP

Ray Williams

April 20, 2013

TRIDENT UNIVERSITY Module 5 - Case Heat and Thermodynamics RAY WILLIAMS 4/24/2013

Module 5 - Case

Heat and Thermodynamics

1. How much energy (in kcal ) is required to raise the temperature of 1 L of water from 20 to 100C, and then turn it to steam at 100C?

VARIABLES:

q(joules)=1000grams

1 liter of water is 1000 milliliters

1L would weigh 1000 grams( ~ 36 oz ) q(joules) = mass * specific heat * (Temp. final - Temp. initial) q = 1000 grams * 4.180 J/gC * (100 C - 20 C) q = 334400 joules

And 1 cal = 4.186 J

So q in kcal, 79885.33 cal = 79.885 kcal

ANSWER: 79.885 kcal

2. A 2 kg rod of aluminum (c = 0.9 kJ/kg-C) at 90C is dropped into 10 L of water at 10C. What is the final temperature of the mixture?

Heat lost by the rod = heat gained by the water

q_lost = q_gain

mass_aluminium * sp.heat_aluminium *(Tf – Ti) = mass_water * sp.heat_water*(Tf-Ti)

1L = 1000 milliliter

1L = 1000g, so 10L = 10000g

2kg* 0.9kJ/kgC * (90C – x) = 10000g * 4.180 J/gC * (x – 10C)

2000g * 0.9 J/gC * (90 – x) = 10000g * 4.180 J/gC * (x – 10C)

1800 (90 –x) = 41800 ( x – 10)

90 –x = 23.222 (x – 10)

90 –x = 23.222x – 232.22

23.222x + x = 232.222 + 90

24.222x = 322.222

X = 13.308 C

ANSWER: 13.308C

3. A 0.2 kg block of an unknown metal at 50C is immersed in 1L of water at 4C. The equilibrium temperature of the mixture is 35C. What is the specific heat of the metal, in kcal/kg-C ?

Heat lost by the rod = heat gained by the water

q_lost = q_gain

mass_metal * sp.heat_metal *(Tf – Ti) = mass_water * sp.heat_water*(Tf-Ti)

1L = 1000 milliliter

1L = 1000g,

0.2kg* x * (50C – 35C) = 1000g * 4.180 J/gC * (35C – 4C)

200g * x* (15C) = 1000g * 4.180 J/gC * (31C)

3000 gC * x = 129580 J

x = 129580 J / 3000 gC

X = 43.1933J/gC

ANSWER: 43.1933 kJ/kgC

4. A liter of gas, initially at a pressure of 500Pa, is compressed from 1.00 L to 0.25 L. During the compression process, heat is dissipated to maintain a constant temperature. What is the final pressure?

Pi = 500 Pa

Vi = 1L

Vf = 0.25L

PV = nRT

PiVi = nRT = 500Pa * 1L

Pf*Vf = nRT = Pf * 0.25L

Equating both,

500Pa * 1L = Pf * 0.25L

Pf = 2000Pa

Final pressure, Pf = 2000 Pa = 2 kPa

ANSWER: 2 kPa

5. A radioisotope thermoelectric generator (RTG) generates electricity by absorbing heat from a core containing a radioactive material (plutonium 238 is typical) and discharging it into the environment. An RTG installed on an unmanned sonobuoy, tethered to the ocean floor, has a core temperature of 150 C, and discharges heat into sea water at 10 C. What is its theoretical efficiency?

T_cold = 10C

T_hot = 150C

Theoretical efficiency = 1 – Tcold / Thot

Theoretical effieciency = 1 – (10/150) = 1-0.0667 = 93.333%

ANSWER: 93.333%