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A piece of gold with a mass of 15.23 g and an initial temperature of 54 °C was dropped into a calorimeter containing 28 g of water. The final temperature of the metal and water in the calorimeter was 62°C. What was the initial temperature of the water?

31.03 °C

62.13 °C

81.03°C

92.13°C

Verified Answer :

MissPhiladelphia

A piece of gold with a mass of 15.23 g and an initial
temperature of 54 °C was dropped into a calorimeter containing 28 g of water.
The final temperature of the metal and water in the calorimeter was 62°C. this
can be solve using the formula H = mCpDeltaT where m is the mass, Cp is the
heat capacity, delta T is the change in temperature

Cp of gold = 0.129 J/ g °C

Cp of water = 4.18 J/ g °C

The H for both are equal , so equate them and solve for the
initial T of water

(15.23g) (0.129 J/ g C) (62 °C -54 °C) = (28 g)( 4.18 J/ g °C)(
Ti –62 °C)

Ti = 62.13 °C

Answer:The initial temperature of the water  62.13 °C.

Explanation:

Heat absorbed by the gold :Q

Mass of the gold ,m= 15.23 g

Specific heat capacity of gold = c = 0.129 J/g °C

Change in temperature of the gold =\(\Delta T\)=62°C-54°C=8°C

Heat lost by the water : Q'

Mass of the water,m' = 28 g

Specif heat of water = c' = 4.179 J/g °C

Change in temperature of the water =\(\Delta T'\)

let the initial temperature of the water be T.

As per as Law of Conservation of Energy:

Q = -Q'

\(m\times c\Delta T=m'\times c'\times \Delta T'\)

\(15.23 g\times 0.129 J/g ^oC\times 8^oC=-(28 g\times 4.179 J/g ^oC\times \Delta T')\)

\(\Delta T'=-0.13^oC\)

\(-(0.13^oC)=(62^oC-T)\)

T = 62.13 °C

The initial temperature of the water  62.13 °C.

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