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a 0.350 kg block at -27.5 °C is added to 0.217 kg of water at 25.0 °C. they come to equilibrium at 16.4 °C. what is the specific heat of the block?

a 0.350 kg block at -27.5 °C is added to 0.217 kg of water at 25.0 °C. they come to-example-1
User Pablito
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2 Answers

3 votes

The specific heat of the block is approximately 0.839 J/g°C.

To find the specific heat of the block, we can use the principle of energy conservation.

First, let's calculate the heat gained by the water:

Q of water = mass of water * specific heat of water * change in temperature of water

Where:
- mass of water = 0.217 kg (given)
- specific heat of water = 4.186 J/g°C (specific heat capacity of water)
- change in temperature of water = equilibrium temperature - initial temperature of water

Plugging in the values, we get:

Q of water = 0.217 kg * 4.186 J/g°C * (16.4°C - 25.0°C)

Next, let's calculate the heat lost by the block:

Q of block = mass of block * specific heat of block * change in temperature of block

Where:
- mass of block = 0.350 kg (given)
- specific heat of block = ? (what we're trying to find)
- change in temperature of block = equilibrium temperature - initial temperature of block

Since the block and water come to equilibrium, the heat lost by the block is equal to the heat gained by the water:

Q of block = Q of water

Now we can set up an equation:

mass of block * specific heat of block * change in temperature_block = Q of water

Plugging in the known values, we get:

0.350 kg * specific heat of block * (16.4°C - (-27.5°C)) = Q of water

Simplifying:

0.350 kg * specific heat_block * 43.9°C = Q_water

Since Q of water = 0.217 kg * 4.186 J/g°C * (16.4°C - 25.0°C), we can substitute this value:

0.350 kg * specific heat of block * 43.9°C = 0.217 kg * 4.186 J/g°C * (16.4°C - 25.0°C)

Simplifying further:

specific heat of block = (0.217 kg * 4.186 J/g°C * (16.4°C - 25.0°C)) / (0.350 kg * 43.9°C)

Calculating this expression, we find the specific heat of the block to be approximately:

specific heat_block ≈ 0.839 J/g°C

User StealthRT
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4.9k points
2 votes

Answer:


C_(pb)=0.501\ kJ/kg.K

Step-by-step explanation:

Given that


m_1=0.35 kg


T_1=-27.5^oC


m_2=0.214 kg


T_2=25^oC


T=16.4^oC

We know that


C_(pw)=4.187 kJ/kg.K

By using energy conservation

Heat lost by water = Heat gain by block


m_2* C_(pw)* (T_2-T)=m_1* C_(pb)* (T-T_1)


0.214* 4.187* (25-16.4)=0.35* C_(pb)* (16.4+27.5)


C_(pb)=0.501\ kJ/kg.K

Therefore the specific heat of the block will be 0.501 kJ/kg.K

User Aruanoc
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