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Zolol [24]
3 years ago
13

In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal.

Chemistry
1 answer:
HACTEHA [7]3 years ago
8 0

Answer : The specific heat of tungsten is, 0.139J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of tungsten = ?

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of tungsten = 19.5 g

m_2 = mass of water = 78.5 g

T_f = final temperature = 23.20^oC

T_1 = initial temperature of tungsten = 97.80^oC

T_2 = initial temperature of water = 22.58^oC

Now put all the given values in the above formula, we get

19.5g\times c_1\times (23.20-97.80)^oC=-78.5g\times 4.18J/g^oC\times (23.20-22.58)^oC

c_1=0.139J/g^oC

Therefore, the specific heat of tungsten is, 0.139J/g^oC

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Answer:

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Explanation:

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3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

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