2. A construction worker drops a hot 100-g iron rivet at 500 °C into a bucket containing 500 g of mercury at 20°C. Assuming that no heat is lost to the surroundings or the bucket, what is the final temperature of the rivet and mercury? 3. An unknown liquid of mass 400 g at a temperature of 80°C is poured into 400 g of water at 40°C. The final equilibrium temperature of the mixture is 49°C. What is the specific heat of the unknown liquid?
Solution
2. Principle of calorimetry states that if there is no loss of heat in surrounding than total heat loss by hot body equals to total heat gained by a cold body.
Now let final tempature of rivet and mercury is T°C
specfic heat of iron rivet C 1 = 448 J/KgC
specfic heat of mercury C 2 = 138 J/kgC
Now using heat transfer take place from a higher tempature object to lower tempature object until an equalibirium tempature is obtained & heat transfer formula when the tempature of an object goes from T 1 to T 2
Q = mC(T 2 -T 1 )
mass of iron rivet m 1 = 100 gm = 0.1 kg,  Intial tempature of iron rivet T 1 = 500°C
mass of mercury m 2 = 500 gm = 0.5 kg, Intial tempature of mercury T 2 = 20°C
Than using Heat Gain = Heat Loss
m 1 C 1 (T 1 -T) = m 2 C 2 (T-T 2 )
(.1kg)(448 J/KgC)(500-T) = (.5 kg)(138 J/kgC)(T-20)
44.8(500 - T) = (69)(T - 20)
22400 - 44.8T = 69T - 1380
2240+1380 = 69T + 44.8T
23780 = 113.8T
T = 208°C
3 .Let the specific heat of unknown liquid is C 1 calorie/gram °C
for liquid
mass m 1 = 400 gm
intial tempature T 1 = 80°C
for water
mass m 2 = 400 gm
intial tempature T 2 = 40°C
specific heat of water  C 2 = 1 calorie/gram °C
Final tempature of solution T = 49°C
Than using Heat Gain = Heat Loss
m 1 C 1 (T 1 -T) = m 2 C 2 (T-T 2 )
(400gm)(C 1 calorie/gram °C)(80°C - 49°C ) = (400 gm)(1 calorie/gram °C)(49°C - 40°C)
C 1 = 0.2903 calorie/gram °C
.
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2- A construction worker drops a hot 100-g iron rivet at 500 C into a.docx
1. 2. A construction worker drops a hot 100-g iron rivet at 500 °C into a bucket containing 500 g
of mercury at 20°C. Assuming that no heat is lost to the surroundings or the bucket, what is the
final temperature of the rivet and mercury? 3. An unknown liquid of mass 400 g at a temperature
of 80°C is poured into 400 g of water at 40°C. The final equilibrium temperature of the
mixture is 49°C. What is the specific heat of the unknown liquid?
Solution
2. Principle of calorimetry states that if there is no loss of heat in surrounding than total heat
loss by hot body equals to total heat gained by a cold body.
Now let final tempature of rivet and mercury is T°C
specfic heat of iron rivet C 1 = 448 J/KgC
specfic heat of mercury C 2 = 138 J/kgC
Now using heat transfer take place from a higher tempature object to lower tempature object
until an equalibirium tempature is obtained & heat transfer formula when the tempature of an
object goes from T 1 to T 2
Q = mC(T 2 -T 1 )
mass of iron rivet m 1 = 100 gm = 0.1 kg,  Intial tempature of iron rivet T 1 = 500°C
mass of mercury m 2 = 500 gm = 0.5 kg, Intial tempature of mercury T 2 = 20°C
Than using Heat Gain = Heat Loss
m 1 C 1 (T 1 -T) = m 2 C 2 (T-T 2 )
(.1kg)(448 J/KgC)(500-T) = (.5 kg)(138 J/kgC)(T-20)
44.8(500 - T) = (69)(T - 20)
22400 - 44.8T = 69T - 1380
2240+1380 = 69T + 44.8T
2. 23780 = 113.8T
T = 208°C
3 .Let the specific heat of unknown liquid is C 1 calorie/gram °C
for liquid
mass m 1 = 400 gm
intial tempature T 1 = 80°C
for water
mass m 2 = 400 gm
intial tempature T 2 = 40°C
specific heat of water  C 2 = 1 calorie/gram °C
Final tempature of solution T = 49°C
Than using Heat Gain = Heat Loss
m 1 C 1 (T 1 -T) = m 2 C 2 (T-T 2 )
(400gm)(C 1 calorie/gram °C)(80°C - 49°C ) = (400 gm)(1 calorie/gram °C)(49°C -
40°C)
C 1 = 0.2903 calorie/gram °C
