A 1.50-kg iron horseshoe initially at 660°C is dropped into a bucket containing 22.5 kg of water at 21.0°C. What is the final temperature of the water-horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away. Step 1 The water will gain the energy lost by the iron. Using the expressions for the quantity of heat exchanged by the two materials, the change in water temperature can be found. The specific heat of iron is 448 J/kg-°C and the specific heat of water is 4,186 J/kg-°C. We assume that the water-horseshoe system is thermally isolated (insulated) from the environment for the short time required for the horseshoe to cool off and the water to warm up. Then the total energy input from the surroundings is zero, as expressed by the equation Qiron + Qwater = 0. The energy Q transferred between a sample of mass m of a material and its surroundings with a temperature change AT is given by Q = mcAT. Substituting the expressions for iron and water into the energy equation, we have (mcAT)iron + (MCAT)water = 0 or, with T as the final temperature of both the iron and the water, miron iron (T- 660 660 °c) + mwater water (T- 21 21 °c) = 0. Note that the first term in this equation results in a negative number of joules, representing energy lost by the originally hot horseshoe, and the second term is a positive number with the same absolute value, representing energy gained in heat by the colder water. Step 2 Solving for the final temperature gives mwater water (21.0°C) + m, T = MironCiron (660°C) mironCiron + mwaterwater ]k9) ] k9)(4,186 J/kg-°c)(21.0°c): 448 J/kg - °C) + (C (448 J/kg - °c)(660°C) |kg)(4186 J/kg • °C) kg °C.
A 1.50-kg iron horseshoe initially at 660°C is dropped into a bucket containing 22.5 kg of water at 21.0°C. What is the final temperature of the water-horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away. Step 1 The water will gain the energy lost by the iron. Using the expressions for the quantity of heat exchanged by the two materials, the change in water temperature can be found. The specific heat of iron is 448 J/kg-°C and the specific heat of water is 4,186 J/kg-°C. We assume that the water-horseshoe system is thermally isolated (insulated) from the environment for the short time required for the horseshoe to cool off and the water to warm up. Then the total energy input from the surroundings is zero, as expressed by the equation Qiron + Qwater = 0. The energy Q transferred between a sample of mass m of a material and its surroundings with a temperature change AT is given by Q = mcAT. Substituting the expressions for iron and water into the energy equation, we have (mcAT)iron + (MCAT)water = 0 or, with T as the final temperature of both the iron and the water, miron iron (T- 660 660 °c) + mwater water (T- 21 21 °c) = 0. Note that the first term in this equation results in a negative number of joules, representing energy lost by the originally hot horseshoe, and the second term is a positive number with the same absolute value, representing energy gained in heat by the colder water. Step 2 Solving for the final temperature gives mwater water (21.0°C) + m, T = MironCiron (660°C) mironCiron + mwaterwater ]k9) ] k9)(4,186 J/kg-°c)(21.0°c): 448 J/kg - °C) + (C (448 J/kg - °c)(660°C) |kg)(4186 J/kg • °C) kg °C.
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Transcribed Image Text:A 1.50-kg iron horseshoe initially at 660°C is dropped into a bucket containing 22.5 kg of water at 21.0°C.
What is the final temperature of the water-horseshoe system? Ignore the heat capacity of the container and
assume a negligible amount of water boils away.
Step 1
The water will gain the energy lost by the iron. Using the expressions for the quantity of heat exchanged by
the two materials, the change in water temperature can be found. The specific heat of iron is 448 J/kg.°C and
the specific heat of water is 4,186 J/kg.°C.
We assume that the water-horseshoe system is thermally isolated (insulated) from the environment for the
short time required for the horseshoe to cool off and the water to warm up. Then the total energy input from
the surroundings is zero, as expressed by the equation
Qiron + Qwater = 0.
The energy Q transferred between a sample of mass m of a material and its surroundings with a temperature
change AT is given by
Q = mcAT.
Substituting the expressions for iron and water into the energy equation, we have
(mcAT)iron + (mcAT),water = 0
or, with T as the final temperature of both the iron and the water,
mironCiron (T- 660
660 °c) + mwater water (T- 21
21 °C
= 0.
Note that the first term in this equation results in a negative number of joules, representing energy lost by the
originally hot horseshoe, and the second term is a positive number with the same absolute value, representing
energy gained in heat by the colder water.
Step 2
Solving for the final temperature gives
Miron iron (660°C)
+ mwater water
mwater water (21.0°C) + m
T =
miron iron
k9)
4,186 J/kg-°C)(21.0°c) +((
/kg • •c) + (C
(448 J/kg · °c)(660°c)
kg
kg
kg)( 448
kg )( 4186 J/kg · °C
°C.
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