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A .35 kg block at -27.5 ºC is added to .217 kg of water at 25.0 ºC. They come to equilibrium at 16.4 ºC. What is the specific heat of the block?

User ChaChaPoly
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Answer:

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To solve this problem, we can use the formula for heat transfer:

q = mcΔT

where q is the heat transferred, m is the mass of the object, c is its specific heat capacity, and ΔT is the change in temperature.

We know that the mass of the block is 0.35 kg and that its initial temperature is -27.5 ºC. We also know that the mass of water is 0.217 kg and that its initial temperature is 25.0 ºC.

When they come to equilibrium at 16.4 ºC, we can calculate how much heat was transferred from the water to the block:

q = mcΔT q = (0.217 kg)(4186 J/kg ºC)(25.0 ºC - 16.4 ºC) q = 1825 J

This amount of heat was transferred from the water to the block, so we can set it equal to the amount of heat absorbed by the block:

q = mcΔT 1825 J = (0.35 kg)c(16.4 ºC - (-27.5 ºC)) 1825 J = (0.35 kg)c(43.9 ºC) c = 148 J/kg ºC

Therefore, the specific heat capacity of the block is 148 J/kg ºC.

Step-by-step explanation:

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User Blitzqwe
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