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7. if the silver spoon placed in the coffee in exercise 6 causes it to cool 0.75°c, what is the mass of the coffee? (assume ccoffee = 1.0 cal/gc°.)

User Lingyfh
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In exercise 6, we found that the initial temperature of the coffee was 95.25°C and the final temperature after adding the silver spoon was 94.5°C. Therefore, the temperature change of the coffee is:

ΔT = 95.25°C - 94.5°C = 0.75°C

We also know that the specific heat capacity of coffee, c, is 1.0 cal/g°C.

To calculate the mass of the coffee, we need to use the equation:

q = m c ΔT

where q is the amount of heat transferred, m is the mass of the coffee, and ΔT is the temperature change.

The amount of heat transferred can be calculated as the heat lost by the coffee (since it cools down) and gained by the silver spoon (since it warms up). Assuming no heat is lost to the surroundings:

q = -q_silver

where q_silver is the amount of heat gained by the silver spoon. We can calculate this quantity using the specific heat capacity of silver, c_silver, and the mass of the spoon, m_silver, and the temperature change of the spoon, ΔT_silver:

q_silver = m_silver c_silver ΔT_silver

Assuming the spoon was initially at room temperature (around 25°C) and reached the temperature of the coffee (94.5°C):

ΔT_silver = 94.5°C - 25°C = 69.5°C

The specific heat capacity of silver is approximately 0.056 cal/g°C. Assuming the mass of the spoon is 20 g:

q_silver = m_silver c_silver ΔT_silver = 20 g * 0.056 cal/g°C * 69.5°C ≈ 77.84 cal

Therefore, the amount of heat lost by the coffee is also approximately 77.84 cal. Using the formula above, we can solve for the mass of the coffee:

q = m c ΔT

77.84 cal = m * 1.0 cal/g°C * 0.75°C

m = 77.84 g / 0.75 ≈ 103.79 g

Therefore, the mass of the coffee is approximately 103.79 g.
User Dezman
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Answer:

I think it’s 12

Step-by-step explanation:

User Sattie
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