A 0.0255 kg silver ring (Cp= 234 J kg-1ºC-1) is heated to 84 ºC and then placed in a calorimeter containing 0.05 kg of water at 24 ºC. The calorimeter is not perfectly insulated, so 0.14 kJ of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?
Answers
Case 2: A 0.780 kg silver pellet with a temperature of 85 °C is added to 0.150 kg of water in a copper cup of 0.66 kg. The initial temperature of the water and the copper cup is 14 °C. Assume no heat is exchanged with the surroundings. The specific heats of silver, water and copper are: 234 J/(kg °C). 4186 J/(kg oC) and 387 J/(kg °C), respectively. Find the final temprature of the system (silver+water+copper cup). Note: 1. Silver pellet releases heat, apply Q = mcAT to silver. To ensure Q is positive, how should you write AT for silver, in terms of the final temperature Tf and silver's intial temperature? (no submission) 2. Water absorbs heat, apply Q = mcAT to water. To ensure Q is positive, how should you write AT for water, in terms of the final temperature Tf and water's intial temperature? (no submission) 3. Copper cup absorbs heat, apply Q-mcAT to copper. To ensure Qis positive, how should you write AT for copper, in terms of the final temperature Tf and copper's intial temperature?? (no submission) 3. Let Q released by silver = Q absorbed by water + Q absorbed by copper, solve for Tf Check that your result of Tf is reasonable. Keep 2 decimal places. (5 attempts remaining) Case 3: A 0.085 kg of an unknown solid sample at a temperature of 100.0°C is dropped into a calorimetor. The calorimetor can is made of 0.56 kg of copper (specific heat = 387 J/(kg °C)) and contains 0.150 kg of water (specific heat = 4186 J/(kg °C), and both the can and water are initially at 14.0°C. The final temprature of the system after thermal equilibrium is established is measured to be 22.5 °C. Assume no heat is exchanged with the surroundings. Calculate the heat absorbed by water. Keep 2 decimal places. Report all heat as POSITIVE e (5 attempts remaining) Calculate the heat absorbed by copper. Keep 2 decimal places. Report all heat as POSITIVE Se (5 attempts remaining) ) of this unknown sample, Keep 1 decimal Find the specific heat (in /(kg place. J/(kg) S (5 attempts remaining)
Please mark me brainliest
Therefore the Final temperature of the water and the silver ring is 25.013 ºC.
Given:
The mass of the silver ring = M = 0.0255 kg
Specific heat of the silver ring = Cp = 234 J/kgºC
Mass of water = m = 0.05 kg
Specific heat of water = C = 4182 J/kgºC
Temperature of the silver ring = T₂ = 84 ºC
Temperature of water = T₁ = 24 ºC
Energy lost to surroundings = E = 0.14 KJ = 140 J
To Find:
The final temperature(T).
Solution:
The given question can be solved as shown below.
From the First Law of Thermodynamics,
⇒ Heat lost = Heat gained
⇒ Heat lost by silver ring = Heat gained by water + heat gained by surroundings
⇒ MCpΔT = mCΔt + E
⇒ [ 0.0255 × 234 × ( 84 - T ) ] = [ 0.05 × 4182 × ( T - 24 ) ] + 140
⇒ 5.967 × ( 84 - T ) = 209.1 × ( T - 24 ) + 140
⇒ 501.228 - 5.967T = 209.1T - 5018.4 +140
⇒ 501.228 + 4878.4 = 209.1T + 5.967T
⇒ 5379.628 = 215.067T
⇒ T = 5379.628/215.067 = 25.013
Therefore the Final temperature of the water and the silver ring is 25.013 ºC.
#SPJ2