== == This answer is taken straight off Yahoo answers and I thought it would be helpful to "spread the wealth." here it is. change in temperature of metal,
75 - 18.3 = 56.7 'C
change in temperature of water,
18.3 - 15 = 3.3 'C
energy gained by water, assuming Cp water = 4.1813 J/g/'C
using the formula, Q = mCp(theta)
where,
Q = energy in Joules
m = mass in grams
Cp = specific heat capacity in J/g/'C
theta = change in temperature in 'C
3.3 * 150 * 4.1813 = 2.06974 kJ
energy gained by water = energy dissipated by metal
using the formula, Q = mCp(theta) and solving for Cp
Cp of metal = 2.06974 k / 56.7 *150 = 0.2434 J/g/'C
Changing 45.2 ml of pure water from 21.00 °C to 23.68 °C requires
45.2 ml x 1g/mL x (23.68-21.00)°C x calorie/g/°C = 121.136 calories
This means the original metal sample had a heat capacity of
121.1.36 cal/58.932 g/(101.00 - 24.68) °C = 0.026585 cal/g/°C = 0.026585 kcal/kg/°C
Possible candidates for the metal if it is pure include:
Uranium (0.028 kcal/kg/°C)
Bismuth (0.03 kcal/kg/°C)
Thallium (0.03 kcal/kg/°C)
Thorium (0.03 kcal/kg/°C)
Note that none of these is quite right, but they are close. A better possibility is that the metal is an alloy.
92.7 degrees Celsius
(Note: Tf is final temp; sw is specific heat of water; mw is mass of water; Tw is initial temp of water; sM is specific heat of metal; mM is mass of metal; and TM is initial temp of metal)
Sounds like a temp. comes to equilibrium and two q's to 0 problem.
(19 g metal)(SH)(22 C - 96 C) + (75 g H2O)(4.180 J/gC)(22 C - 18 C) = 0
-1406(SH) + 1254 = 0
-1406(SH) = -1254
specific heat of metal = 0.89 J/gC
Two q's to 0. q eliminated algebraically, so....
mass * specific heat * change in temperature + mass * specific heat * change in temperature = 0
(12 grams metal)(X specific heat)(25o C - 80o C) + (50 grams H2O)(4.180 J/gC)(25o C - 10o C) = 0
(- 660)X + (3135) = 0
- 660X = - 3135
X = 4.8 J/gC
===========specific heat metal
Heat loss by metal = heat gain by water (Sp. heat )x mass x temp change of metal = (Sp. heat) x mass x temp change of water. c x 50 x (100-22) = 1 x 400 x (22-20)3900 c = 800c = 800/3900 = 8/39 approximately 0.3 cal/(g-deg C).
Lead, copper, Aluminum, or iron
that was the answer choices :p
This was my question and i ddnt have enough room to right the rest... my answer was 0.88 J/g c but i am not confident in my answer. does anyone know how to do this or if my answer is right?
20.238 degrees C (to 3 d.p.)
Ok
1 degree Celsius = 33.8 degrees Fahrenheit 2 degrees Celsius = 35.6 degrees Fahrenheit 35.6 - 33.8 = 1.8
A change of 7.2 degrees on the Fahrenheit scale would be a change of 4 degrees on the Celsius scale. The conversion is 1.8 Fahrenheit degrees per Celsius degree (Celsius degrees are larger intervals).Example :[ Temperature in °C = 5/9 (temperature in °F - 32°) ]86° F is equal to 30° C93.2° F is equal to 34° C
No, the "degrees" have the same name but are different sizes. (Celsius degrees are larger intervals than Fahrenheit degrees.) A change of 1 degree Celsius is the same as a change of 1.8 degrees Fahrenheit.
39 degrees Celsius = 102.2 degrees Fahrenheit.
13.16 kJ
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Quantity of Energy= mass x temperature change x specific heat capacity For example: Find the amount of energy needed to raise the temperature of 0.20 kg of lead by 15 degree Celsius if the specific heat capacity of lead is 0.90 J/g degree Celsius. Answer: J=200g x 15 degree Celsius x 0.90 J/g degree Celsius = 2700 J
Is the amount of energy that is required to change the temperature of 1kg of a substance by 1 degree Celsius with no state change.
heat energy required to raise the temperature of ice by 29 celsius =specific heat capacity of ice * temperature change *mass of ice + to change 1kg of ice at 0 celsius to water at 0 celsius =specific latent of fusion of ice*mass of water + heat energy required to raise the temperature of water by 106 celsius =specific heat capacity of water * temperature change *mass of ice + to change 1kg of water at 106 celsius to steam at 106 celsius =specific latent of fusion of ice*mass of steam
20 degrees Celsius is 68 degrees Fahrenheit. 45 degrees Celsius is 113 degrees Fahrenheit. So 25 degrees change Celsius = 25 x 1.8 = 45 degrees Fahrenheit. 1 degree change in Celsius is equivalent to 1.8 degrees change in Fahrenheit. Scroll down to related links and look at "Conversion of Temperature Units".
59 degrees Fahrenheit = 15 degrees Celsius
46 degrees Fahrenheit = 7.78 degrees Celsius
The change is 7 Celsius degrees (NOT degrees Celsius).
The equation for specific heat is: C = q/temp. change x mass. C is a substance's specific heat, which is a constant for every substance. q is its heat capacity in joules, temp. change is the change in temperature in degrees Celsius, and mass is in grams.
1 degree Celsius = 33.8 degrees Fahrenheit 2 degrees Celsius = 35.6 degrees Fahrenheit 35.6 - 33.8 = 1.8
Two q's to 0. q = mass * specific heat * change in temp.. Q falls out of equation(50 g)(X specific heat)(30 C - 80 C) + (100 g H2O)(4.180 J/gC)(30 C - 25 C) = 0(- 2500X) + (2090) = 0- 2500X = - 2090X = 0.836 J/gC=============the specific heat of the solid
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