Question

QUESTION 3 The specific heat (s) of a substance is the amount of heat (9) required to raise the temperature of one gram of the substance by one degree Celsius. Calculation of heat (q) absorbed or released: q=sxm x At where At = tfinal - tinitial TABLE 6.2 The Specific Heats of Some Common Substances Specific Heat Substance (J/g. "C) AI 0.900 Au 0.129 C (graphite) 0.720 C (diamond) 0.502 Cu 0.385 Fe Hg 0.139 HO 4.184 C,H,OH (ethanol) 2.46 QUESTION A 100 g iron bar at 100 °C was placed in a beaker that contains 1000 g of water at 10 °C. What is the final temperature of the water. No heat is assumed to be released. The beaker is assumed to be independent of heat transfer. 0.444 gits the decimal point. For your answer, type in only the number that carries two Do not type in the unit that is °C. Hint: Refer to the analogy in Question #2. But in this question, only two family members (mom & dad or daughter & dad) are involved in the transaction. Copyright © All Rights Reserved by M.Han

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Answer 1

Answer -

Given,

Mass of Iron Bar = 100 g

Initial Temperature of Iron Bar = 100 ^$\degree$ C

Mass of water = 1000 g

Initial Temperature of Water = 10 ^$\degree$ C

Specific Heat of Iron = 0.444 J/g^$\degree$C

Specific Heat of Water = 4.184 J/g^$\degree$C

Final Temperature of Water = ?

We know that,

q = s * m * (t_final - t_initial)

where,

q = Heat

s = specific Heat

t = Temperature

Also,

Heat released by Iron bar = - Heat absorbed by Water

So,

q_iron = - q_water

s_iron * m_iron * (t_final-iron - t_initial-iron) = -[s_water * m_water * (t_final-water - t_{initial-water})]

Also,

Final Temperature of Iron = Final Temperature of Water

So,

s_iron * m_iron * (t_final-water - t_initial-iron) = -[s_water * m_water * (t_final-water - t_{initial-water})]

s_iron * m_iron * (t_final-water - t_initial-iron) = s_water * m_water * ( t_{initial-water}-t_final-water)

Put the values,

0.444 J/g^$\degree$C * 100 g * (t_final-water - 100 ^$\degree$ C) = 4.184 J/g^$\degree$C * 1000 g * ( 10 ^$\degree$ C - t_final-water)

(0.444 J/g^$\degree$C * 100 g * t_final-water ) - (0.444 J/g^$\degree$C * 100 g * 100 ^$\degree$ C) = (4.184 J/g^$\degree$C * 1000 g * 10 ^$\degree$ C) - (4.184 J/g^$\degree$C * 1000 g * t_final-water)

(44.4 J/^$\degree$C * t_final-water ) - (4440 J) = (41840 J) - (4184 J/^$\degree$C * t_final-water)

(44.4 J/^$\degree$C * t_final-water ) + (4184 J/^$\degree$C * t_final-water) = (41840 J) + (4440 J)

t_final-water * (44.4 J/^$\degree$C + 4184 J/^$\degree$C) = (41840 J + 4440 J)

t_final-water = (41840 J + 4440 J) / (44.4 J/^$\degree$C + 4184 J/^$\degree$C)

t_final-water = 10.94 ^$\degree$ C

Now,

°C + 273.15 = K

So,

10.94 °C + 273.15 = 284.09 K

So, Final Temperature of Water = 284.09 K