Question

You are at a restaurant eating lunch with a friend. After the meal, your friend realizes they left their wallet at home. Your friend gives you a piece of silver with a mass of 0.11 kg as payment for their portion of the meal, which is certainly fair compensation if the metal is actually pure silver. When you get home, you begin to question whether or not you should remain friends.

To test the piece of metal, you boil it in water until it reaches a temperature of 100°C. You also prepare a lead container of mass 1.1 kg and pour 0.64 kg of water into it, both at the room temperature of 19.9°C. The lead container and its contents are well isolated from their surroundings. You remove the piece of metal from the boiling water and place it in the water in the lead container. The entire “silver”-­lead­water system reaches a final temperature of 21.5°C.

Find the specific heat, in units of J/(kg⋅°C), of the metal your friend gave you. Since the specific heat of silver is 235 J/(kg⋅°C) and the specific heat of lead is 130 J/(kg⋅°C), you can judge for yourself whether you are still friends.

Answer 1

To find the specific heat of the metal your friend gave you, we can use the principle of conservation of energy. The heat gained by the metal is equal to the heat lost by the water and lead container. We can calculate the heat gained by the metal using the equation:

Q = m * c * ΔT

Where:

Q is the heat gained or lost,

m is the mass of the object,

c is the specific heat of the substance, and

ΔT is the change in temperature.

Let's calculate the heat gained by the metal:

Heat gained by the metal = Heat lost by the water + Heat lost by the lead container

m * c * ΔT (metal) = (m_water * c_water * ΔT_water) + (m_lead * c_lead * ΔT_lead)

Given:

Mass of the metal (m) = 0.11 kg

Specific heat of silver (c_silver) = 235 J/(kg⋅°C)

Specific heat of lead (c_lead) = 130 J/(kg⋅°C)

Initial temperature of the metal (T_initial) = 100°C

Final temperature of the system (T_final) = 21.5°C

Mass of the water (m_water) = 0.64 kg

Mass of the lead container (m_lead) = 1.1 kg

Initial temperature of the water and lead container (T_water_lead_initial) = 19.9°C

ΔT (metal) = T_final - T_initial

ΔT_water = T_final - T_water_lead_initial

ΔT_lead = T_final - T_water_lead_initial

Let's substitute the values into the equation and solve for the specific heat of the metal:

(0.11 kg) * c * (21.5°C - 100°C) = (0.64 kg) * (4186 J/(kg⋅°C)) * (21.5°C - 19.9°C) + (1.1 kg) * (130 J/(kg⋅°C)) * (21.5°C - 19.9°C)

Simplifying the equation:

-79.35 * c = 1089.44 + 26.4

-79.35 * c = 1115.84

c ≈ -14.06 J/(kg⋅°C)

The specific heat of the metal, calculated as approximately -14.06 J/(kg⋅°C), is not consistent with the specific heat of silver (235 J/(kg⋅°C)). Therefore, it suggests that the metal your friend gave you is not pure silver. Based on this information, you can make your own judgment regarding your friendship.

Step-by-step explanation:

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Answer 2

Comparing the calculated specific heat (280 J/(kg·°C)) to the known specific heat of silver (235 J/(kg·°C)) indicates the metal isn't pure silver, prompting potential discussions with your friend.

To figure out the specific heat of the metal your friend provided, apply the principle of conservation of energy. The heat gained by the cooler substances (lead and water) equals the heat lost by the hotter substance (the metal). The heat transfer formula is Q = mcΔT, where Q is heat, m is mass, c is specific heat, and ΔT is the change in temperature.

Let m1 represent the mass of the metal, c1 its specific heat, and ΔT1 its temperature change. Similarly, let m2, c2, and ΔT2 represent the mass, specific heat, and temperature change for the lead and water system.

The heat gained by the lead-water system equals the heat lost by the metal:

m1c1ΔT1 = m2c2ΔT2

Solving for c1:

c1 = (m2c2ΔT2) / (m1ΔT1)

Substitute known values:

c1 = (1.1 kg × 130 J/(kg·°C) × (21.5 - 19.9) °C) / (0.11 kg × (100 - 21.5) °C)

After calculation, c1 is approximately 280 J/(kg·°C).

Comparing this value to the specific heat of silver (235 J/(kg·°C)) suggests the metal isn't pure silver. Your friend might have unknowingly given you an alloy or a different material. However, it's crucial to communicate openly before making conclusions about the friendship.