Specific heat capacity?

Question

A 150.0 g sample of a metal at 75.0'C is added to 150.0g of H2O at 15'C. The temperature of the water rises to 18.3'C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

Answer 1

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

Heat Capacity Of Metal

Answer 3

This Site Might Help You.

RE:
Specific heat capacity?
A 150.0 g sample of a metal at 75.0'C is added to 150.0g of H2O at 15'C. The temperature of the water rises to 18.3'C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

Answer 4

specific heat capacity

Answer 5

heat gained by water = (1 cal/(g degree)) (150 g) (18.3-15 degrees)

Heat lost = C m delta T

Solve for C then substitute values. Keep in mind that the heat lost by metal = heat gained by water and the final temperature of both is the same.

Answer 6

For any question like this, you only have one equation:

-mcDT=mcDT

where m are the masses of the two substance, c are their specific heats and DT are the temperature changes each experienced.

In any of these problems, you'll be given everything except one of the values. You should be able to calculate that one.

Answer 7

Hello!
Q=mc∆T

The best way is to organize your given data:

(metal) mc∆T=mc∆T (water)

m(water) 150g (always convert to kg i.e. 0.150kg
∆T (water) = 75'C (always convert to kelvin i.e. 273+75 =348)
c (water) = 4200 J
____________________________________________

m(metal) = 0.150kg
∆T (metal) = 18.3 + 273 = 291.3 K
c (metal) = unknown

Back to our formula:
(metal) mc∆T=mc∆T (water)
energy in = energy out

Use your skills of subject making to solve for c (metal)

c(metal) = (water) mc∆T/mc (metal)
c = 0.150(4.2)(348)/ 0.150(291.3)
c = you can calculate

note: please taken thee 15'C into consideration
thank you

Answer 8

sound,s like you allready know the answer to this