A 275-g sample of nickel at 100.0°C is placed in 100.0 g of water at 22.0°C. What is the final temperature of the water? Assume no heat transfer with the surroundings. The specific heat of nickel is 0.444 J/g·°C and the specific heat of water is 4.184 J/g·°C. Hint: The final temp for both the system and surroundings will be the same.
A 275-g sample of nickel at 100.0°C is placed in 100.0 g of water at 22.0°C....
QUESTION 6 A 275-9 sample of nickel metal is placed in 1000 g of water at 220 °C. Ater some time, the temperature of the water reaches 27 5°C What is the initial temperature of the nickel metal? Assume that no heat is lost toorged to the surroundings Specificheat capacity of nickel 0444 C ) * 294 "C 08240 403 od 7020 6100
A 19 g sample of an alloy at 98.0°C is placed into 84.6 g of water at 22.0 °C in an insulated coffee cup with a heat capacity of 9.2 J/K. If the final temperature of the system is 35.0°C, what is the specific heat capacity of the alloy in J/(g.K)? Don't include units. cH2O = 4.184 J/g.K
A calorimeter contained 79.0 g of water at 15.75°C. A 120.-g sample of iron at 63.82°C was placed in it, giving a final temperature of 19.06°C for the system. Calculate the heat capacity of the calorimeter. Specific heats are 4.184 J/g·°C for and 0.444 J/g·°C for . Heat capacity of the calorimeter = ________J/°C
A metal sample weighing 43.5 g at a temperature of 100.0 °C was placed in 39.9 g of water in a calorimeter at 25.1°C. At equilibrium, the temperature of the water and metal was 33.5°C. What was ΔT for the water? (ΔT = Tfinal - Tinitial) What was ΔT for the metal? Using the specific heat of water (4.184 J/g°C), calculate how much heat flowed into the water. Calculate the specific heat of the metal.
A piece of metal of mass 35.0 g at 100.0°C was placed in 150.0 g of water at 20.0 °C. After stirring, the final temperature of the water and the metal is 23.8°C. What is the specific heat capacity of the metal? (specific heat capacity for H2O = 4.184 J/g °C) O-0.89 J 8°C 19.6 J/g °C 1.96J/g °C O 0.89 J/g °C
A 10.95 g sample of lead at 88.0°C was placed into a styrofoam cup calorimeter which contained 15 mL of water at 22.0°C. The final temperature in the calorimeter reached 23.5°C. Calculate the specific heat of lead. The specific heat of water is 4.184 J/g°C.
A piece of titanium at 100.0°C was dropped into 50.0 g of water at 20.0°C. The final temperature of the system was 22.6°C. What was the mass of the titanium? Specific heat (J/g°C) titanium 0.54 water 4.184
a 312 g sample of a metal is heated to 355.272 c A 312 g sample of a metal is heated to 355.272 °C and plunged into 200 g of water at a temperature of 45.471 °C. The final temperature of the water is 59.19 °C. Assuming water has a specific heat capacity of 4.184 J/g °C, what is the specific heat capacity of the metal sample, in J/g °C)? Assume no heat loss to the surroundings. Report your response...
A metal sample weighing 42.6 g and at a temperature of 100.0 oC was placed in 40.6 g of water at 24.8oC. At equilibrium the final temperature of the water and metal was 35.0oC. a. What was the change in temperature for the water? oC b. What was the temperature change for the metal? oC c. Taking the specific heat of water to be 4.184J/goC,caculate the specific heat of the metal. J/goC d. What is the approximate molar mass of...
. A 150.0 g sample of a Metal was heated to 95.0°C. When the hot metal was placed into 100.0 g of water in a calorimeter, the temperature of the water increased from 20.0°C to 35.0°C. The specific heat of water is 4.184 J/g °C. a) What is the specific heat of the metal? Kb) What would the final temperature be if the mass of water was 150.0 q?