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How much heat, in joules, must be added to a 5.00×10 2 g iron skillet to increase its temperature from 25°C to 250 °C? The specific heat of iron is 0.451 J/g °C.

Short Answer

Expert verified

The amount of heat, in joules, is \({\rm{5}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{J}}\).

Step by step solution

01

Given information

Specific heat (c) of the iron =0.451 J/g °C.

Weight of the iron skillet = 5.00×10 2-g = 500g.

From the specific heat of the iron, it is clear that 0.451 g of heat is required to heat 1g of iron by 1°C.

By using the equation below, we can calculate heat in joules.

\(\begin{array}{c}{\bf{q = c \times m \times \Delta T}}\\{\bf{ = c \times m \times (}}{{\bf{T}}_{{\bf{final}}}}{\bf{ - }}{{\bf{T}}_{{\bf{initial}}}}{\bf{)}}\end{array}\)

02

Calculations

The known values (Step 1) are substituted as shown below.

\(\begin{array}{c}{\rm{q = c \times m \times (}}{{\rm{T}}_{{\rm{final}}}}{\rm{ - }}{{\rm{T}}_{{\rm{initial}}}}{\rm{)}}\\\begin{array}{*{20}{l}}{{\rm{ = 0}}{\rm{.451 J/g ^\circ C \times 500g \times }}\left( {{\rm{250 ^\circ C - \;25^\circ C}}} \right)}\\{{\rm{ = 0}}{\rm{.451 J\; \times 500 \times }}\left( {{\rm{225}}} \right)}\\\begin{array}{l}{\rm{ = \;0}}{\rm{.451 \times 112500 }}\\{\rm{ = 50, 737}}{\rm{.5 J }}\\{\rm{ = 5}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{J}}\end{array}\end{array}\end{array}\)

Thus, the required amount of heat, in Joules, is \({\rm{5}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{J}}\).

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Most popular questions from this chapter

Question 11: A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.

(a) What is the specific heat of the substance?

(b) If it is one of the substances found in Table 5.1, what is its likely identity?

Water gas, a mixture of \({{\bf{H}}_{\bf{2}}}\) and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon:\({\bf{C}}\left( {\bf{s}} \right){\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right) \to {\bf{CO}}\left( {\bf{g}} \right){\bf{ + }}{{\bf{H}}_{\bf{2}}}\left( {\bf{g}} \right)\).

(a) Assuming that coke has the same enthalpy of formation as graphite, calculate \({\bf{\Delta H}}_{{\bf{298}}}^{\bf{0}}\)for this reaction.

(b) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst:\({\bf{2}}{{\bf{H}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{ + CO}}\left( {\bf{g}} \right) \to {\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\left( {\bf{g}} \right)\).

Under the conditions of the reaction, methanol forms as a gas. Calculate \({\bf{\Delta H}}_{{\bf{298}}}^{\bf{0}}\)for this reaction and for the condensation of gaseous methanol to liquid methanol.

(c) Calculate the heat of combustion of 1 mole of liquid methanol to H2O(g) and CO2(g).

When 50.0g of 0.200M NaCl(aq) at 24.1˚C is added to 100.0g of 0.100M AgNO3(aq) at 24.1˚C in a calorimeter, the temperature rises to 25.2˚C as AgCl(s) forms. Assuming the specific heat of the solution and products is 4.20J/g˚C, calculate the approximate amount of heat in joules produced.

Question:How much heat, in joules, must be added to a 5.00×102-g iron skillet to increase its temperature from 25°C to 250 °C? The specific heat of iron is 0.451 J/g °C.

Aluminum chloride can be formed from its elements:

(i)\({\bf{2Al(s) + 3C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s) \Delta H^\circ = ?}}\)

Use the reactions here to determine the ΔH° for reaction(i):

\(\begin{array}{*{20}{l}}{\left( {{\bf{ii}}} \right){\rm{ }}{\bf{HCl(g)}} \to {\bf{HCl(aq) \Delta H^\circ (ii) = - 74}}{\bf{.8 kJ}}}\\{\left( {{\bf{iii}}} \right){\rm{ }}{{\bf{H}}_{\bf{2}}}{\bf{(g) + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2HCl(g) \Delta H^\circ (iii) = - 185 kJ}}}\\{\left( {{\bf{iv}}} \right){\rm{ }}{\bf{AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(aq)}} \to {\bf{AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(s) \Delta H^\circ (iv) = + 323 kJ}}}\\{\left( {\bf{v}} \right){\rm{ }}{\bf{2Al(s) + 6HCl(aq)}} \to {\bf{2AlC}}{{\bf{l}}_{\bf{3}}}{\bf{(aq) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g) \Delta H^\circ (v) = - 1049 kJ}}}\end{array}\)

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