Question

Calculate the amount of heat added to 1 gram gold to change phase from solid to liquid. The heat of fusion for gold is 64.5 x 103 J/kg.​

Answer 1

Answer:

Q = m LF

Q = (1 x 10-3 kg)(64.5 x 103 J/kg)

Q = 64.5 Joule

Step-by-step explanation:

Mass (m) = 1 gram = 1 x 10-3 kg

Heat of fusion (LF) = 11.8 x 103 J/kg

Wanted : Heat (Q)

Solution :

Q = m LF

Q = (1 x 10-3 kg)(11.8 x 103 J/kg)

Q = 11.8 Joule

3. Determine the amount of heat absorbed by 1 kg water to change phase from liquid to vapor (steam). Heat of vaporization for water = 2256 x 103 J/kg

Known :

Mass (m) = 1 kg

Heat of vaporization (LV) = 2256 x 103 J/kg

Wanted : Heat (Q)

Solution :

Q = m LV

Q = (1 kg)(2256 x 103 J/kg)

Q = 2256 x 103 Joule

Mass (m) = 1 gram = 1 x 10-3 kg

Heat of vaporization (LV) = 200 x 103 J/kg

Known : Heat (Q)

Solution :

Q = m LV

Q = (1 x 10-3 kg)(200 x 103 J/kg)

Q = 200 Joule

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Answer 2

Given: 1 gram gold changes phase from solid to liquid. The heat of fusion for gold is 64.5 x 10³ J/kg.​

To find: The amount of heat added to the 1 gram gold.

Solution:

Latent heat of fusion of a substance is the amount of heat energy required to change the state of the substance from solid to liquid. It is given by the following formula.

$Q = mL$

Here, Q is the latent heat of fusion, m is the mass and L is the specific heat of fusion of gold.

The mass of the gold is given as 1 gram which when converted to kg is equal to 0.001 kg.

$Q = 0.001 * 64.5 * 10^{3}$

   $= 64.5 J$

Therefore, the amount of heat added to the 1 gram gold is 64.5 J.