1. Calculate the final temperature of the system: A 50.0 gram sample of water in
ID: 964235 • Letter: 1
Question
1. Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 4.57 °C are mixed. The specific heat of water is 4.184 J/gC). Record your answer in scientific notation using 3 significant figures.
2. And unknown sample is being evaluated in lab. What is the specific heat capacity of the compound if it requires 141.19 J to raise the temperature of 80.96 grams of the unknown from 15.2 °C to 78.42 °C. Record your answer to 3 decimal spaces.
3. A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 13.21 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C). Record your answer in scientific notation using three significant figures.
4. How much heat (in joules) is used to heat a 30.86 gram sample of iron from 0.63 degrees Celcius to 100.00 degrees Celcius if the specific heat of Fe is 0.450 j/g*C? Record your answer to 2 decimal spaces.
5. You order a 16 oz glass of tea (where the mass of water is 474 grams) from a local restaurant. The tea is freshly brewed and has an initial temperature of 21.12 °C. You add ice to cool it. If the heat of fusion of ice is 6.020 kJ/mol and each ice cube contains exactly 1 mol of water, how many ice cubes are necessary to cool the tea to 0.96 °C? The specific heat of the "tea" is 4.184 J/g*C. Record your answer as a whole number.
6. A 65.77 gram sample of iron (with a heat capacity of 0.450 J/g ºC) is heated to 100.0 ºC. It is then transferred to a coffee cup calorimeter containing 34.64 g of water (specific heat of 4.184 J/ g ºC) initially at 20.63 ºC. If the final temperature of the system is 23.59, what was the heat capacity of the calorimeter? Record your answer as a whole number (assume the sign is positive).
Explanation / Answer
1. heat lost by hot water = heat gained by water
m*s*DT = m*s*DT
50*4.18*(100 - x) = 100*4.18*(x - 4.57)
x = final temperature = 36.38 C
2. q = m*s*DT
141.19 = 80.96*x*(78.42-15.2)
x = specific heat of sample = 0.0276 j/g.c
3.
25*x*(99-20.15) = 50*4.184*(20.15-13.21)
x = specific heat of metal = 0.736 j/g.c
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