A metal sample weighing 147.90g and at a temperature of 99.5 degree C was placed

Question

A metal sample weighing 147.90g and at a temperature of 99.5 degree C was placed in 49.73 g of water in a calorimeter at 23.0 dgereeC . at equilibrium the temperature of the water and metal was 41.8 degree C what was delta t for the water?(delta t=t_final-t_initial) degree C what was the delta t for the metal? degree how much heat flowed into the water?(Take the specific heat of the water to be 4.18 j/g degree C) joules calculate the specific heat of the metal, using Equation 3 joules/g ^degree C what is the approximate molar mass of the metal?(use Eq.4.) g/mol when 4.98 g of NaOH was dissolved in 49.72g of water n calorimeter at 23.7 degree C. the temperature solution of the solution went up to 50.1 degree C is this dissolution reaction exothermic? Why? Calculate q_h_2o using equation 1. Joules Find delta H for the reaction as it occurred in the calorimeter delta H= joules

Explanation / Answer

Calorimetry is the study of heat flow from one substance to another. A calorimeter is an insulated container that allows heat flow between substances, but does not allow heat to escape.

q = m·s·T

q = heat in Joules (J)

m = mass in grams (g)

T = temperature change = Tfinal - Tinitial

Cs = specific heat = heat required to raise the temp. of 1 g of substance by 1 o C. Units are J/g·o C.

We will essentially put hot metal into cold water. Heat flows from the metal to the water. Mathematically, this process is illustrated by:

qmetal = - qwater

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1. Given data:

a. T of water

T = 41.8°C - 23°C = 18.8°C

b. T of Metal

T = 41.8°C - 99.5°C = -57.7°C

c. Heat gained into water

qwater = mwater·swater·Twater

q water = (49.73 g)(4.18 J/g·°C)(18.8°C)

qwater = 3.908 x 103 J

d. Specific heat of Metal

q metal = - qwater

qmetal = -3.908 x 103 J

Cs = qmetal / m(metal) x T = -3.908x103 J/ (147.90 g)(-57.7°C)

Cs = 0.56 J/g.°C

e. Molar mass Metal

(more information needed)

I could guess;

Atomic Mass ( 4.18 J/mol°C)/(0.5602 J/g °C) = 7.4 g/mol or Li - Not sure!

2. Given Data:

a. Yes, the solution is exothermic because heat is released by the system to warm the surroundings.

b. Heat of water

qwater = mwater·swater·Twater

q water = (49.72 g)(4.18 J/g·°C)(50.1°C - 23,7°C)

qwater = 5.486 x 103 J

c. H for Reaction

H = qwater

H = 5.486 x 103 J

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