Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The specific heat of a metal is determined by heating the weighed metal, adding

ID: 809864 • Letter: T

Question

The specific heat of a metal is determined by heating the weighed metal, adding it to water in a calorimeter, and measuring the temperature change. The specific heat of water is 4.180 Joules per g per C. The data collected are as follows:

Grams of water in the calorimeter 45.14
Grams of metal 19.01
Initial temperature of metal 99.56
Initial temperature of water (C) 22.14
Maximum temperature of water after metal is added (C) 23.59

Calculate the following:
(1) Temperature fall of metal

(2) Temperature rise of water

(3) Joules gained by water (= joules lost by metal)

(4) Specific heat of metal

____________________________________________________

The heat of solution is found by adding a salt to water in a calorimeter and measuring the temperature change.

The specific heat of water is 4.180 Joules per g per C. In the calculation of the heat of solution, ignore the contribution to specific heat and mass due to the salt. Assume that these contributions are negligible.

The data collected are as follows:

Grams of water in the calorimeter 49.58
Grams of salt 4.4301
Initial temperature of water 22.42
Final Temperature 18.92

Calculate the following:

(5) Change in water temperature (C)

(6) Change in Joules in water

(7) Heat of solution of salt

Explanation / Answer

Temperature fall of metal

It will be equal to the temperature rise in water = 23.59 - 22.14oC = 1.45oC

Temperature rise of water

It will same as the final temp - initial temp =  23.59 - 22.14oC = 1.45oC

Joules gained by water (= joules lost by metal)

we know, Q = m x Cp x delta T

delta T = 1.45oC

m = mass of water = 45.14g

Cp = 4.18J/g/oC

So Q = 45.14 x 4.18 x 1.45 = 273.59J

Specific heat of metal

use Q = m x Cp x delta T

delta T = 1.45oC

m= mass of metal = 19.01g

Q = 273.59 J

So , Cp = Q/(mx delta T) = 273.59/ ( 19.01 x 1.45oC) = 9.925J/g/oC

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote