What is the specific heat capacity of a 50 gram piece of 100 degrees celsius metal that will change 400 grams of 20 degrees celsius water to 22 degrees celsius?

Answer 1

== == 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

Answer 2

First, calculate the heat energy gained by the water: Q = mcΔT, where m = mass of water, c = specific heat capacity of water, and ΔT = change in temperature. Next, calculate the heat energy lost by the metal using the same formula but for the metal piece. The heat lost by the metal = heat gained by the water. Use this equality to find the specific heat capacity of the metal.

Answer 3

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.

Answer 4

92.7 degrees Celsius

• Since the heat gained by the water equals the heat lost by the metal, we can equate the two equations for specific heat together(and remember the second equation is negative because it is heat lost): swmw(Tf-Tw) = -sMmM(Tf-TM)

(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)

• We can solve for Tf to get: Tf = (swmwTw + sMmMTM)/(swmw + sMmM)
• After plugging in 4.186 for specific heat of water, 50 for mass of water, 20 for initial temp of water, 2.09 for specific heat of metal, 1000 for mass of metal(convert kg to g), and 100 for initial temp of metal, you'll get Final temp equals 92.7 degrees.
• This is a reasonable answer because the mass of metal is way larger than the mass of water, therefore, making the final temp closer to the initial temp of the metal.

Answer 5

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

Answer 6

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

Answer 7

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).

Answer 8

Lead, copper, Aluminum, or iron

that was the answer choices :p

Answer 9

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?

Answer 10

20.238 degrees C (to 3 d.p.)