Question

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?

Answer 1

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 C₁ = 448 J/KgC

specfic heat of mercury C₂ = 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 T₁ to T₂

                                     Q = mC(T₂-T₁)

mass of iron rivet m₁= 100 gm = 0.1 kg,   Intial tempature of iron rivet T₁ = 500°C

mass of mercury m₂= 500 gm = 0.5 kg, Intial tempature of mercury T₂ = 20°C

Than using Heat Gain = Heat Loss

                m₁C₁(T₁-T) = m₂C₂(T-T₂)

               (.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 C₁ calorie/gram °C

for liquid

mass m₁ = 400 gm

intial tempature T₁ = 80°C

for water

mass m₂ = 400 gm

intial tempature T₂ = 40°C

specific heat of water   C₂= 1 calorie/gram °C

Final tempature of solution T = 49°C

Than using Heat Gain = Heat Loss

                m₁C₁(T₁-T) = m₂C₂(T-T₂)

                (400gm)(C₁ calorie/gram °C)(80°C - 49°C ) = (400 gm)(1 calorie/gram °C)(49°C - 40°C)

               C₁ = 0.2903 calorie/gram °C