(a) Temperature of 30 g of water at 22 C rises to 31.5 C by dipping a hot copper
ID: 700621 • Letter: #
Question
(a) Temperature of 30 g of water at 22 C rises to 31.5 C by dipping a hot copper metal piece. The total rise in temperature of water is 9.5 C (31.5 C -22 C). Use specific heat of water to calculate the energy required to raise the temperature of water.
Specific heat of water is the energy required by one gram of water per degree rise in temperature. It is given as 4.184J/(g. C) for water.
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Using this value calculate the amount of heat required to raise the temperature of water.
Energy (E1) required will be = Specific heat x Amount of water x C rise in temperature
= 4.184J/(g. C) x 30 g x 9.5 C
= 1192.44 J
=1.19 x 103J or 1.2 x 103J
(b) The temperature of copper metal piece drops from 100 C to 31.5 C i.e. a drop in temperature of 68.5 C (100 – 31.5) C.
When this hot copper metal is dipped in water, amount of heat lost by the copper metal piece is equal to heat gained by 30 g of water.
Heat gained by water is calculated in the previous part of this question i.e. 1192.44 J.
Now calculate specific heat of copper metal by using the above data.
Let us assume that specific heat of copper is . Amount of heat required to raise the temperature of 46.0 g of copper by 61.5 degree centrigrade is the same as the heat lost by the metal in bringing down its temperature by 61.5 degree centrigrade. This is calculated in previous part of this question.
Energy required = Specific heat x Amount of copper x C rise in temperature
1192.44 J = x 46 g x 68.5 C
0.378 J/(g. C) =
Explanation / Answer
4) A 46.0-g sample of copper is heated to 100.0 °C and the hot metal is quickly dropperd into 30.0 g of cold water in a Styrofoam cup calorimeter. The temperature of water and calorimeter was initially at 22.0 °C and the final temperature of water calorimeter, and copper is 31.5 °C, (a) Calculate the total amount of heat absorbed by water and calorimeter. (b) Calculate the specific heat of copper. (Specific heat of water is 4.184 Jig. °C. Assume that heat absorbed by calorimeter is negligible.) (Ans: (a) heat absorbed by water, q 1.2 x 10 J; (b) specific heat of copper 0,38 Jlg. °c) 5) 50.0 g of aluminum at 98.0 °C is added 40.0 g of water in a Styrofoam cup calorimeter. If the initial temperature of water were 22.0 °C, what is the final temperature of water and aluminum? (Specific heat of water is 4.184 J/g. °C and specific heat of aluminum is 0.89 J/g. °C. Assume that heat absorbed by calorimeter is negligible.) (Ans: final temperature, T-38.0 °C) 6) Two gases, A(g) and B(g), are confined in a cylinder-and-piston arrangement. Substances A and B react to form a solid product A(g) + B(g) C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? (Ans: -670 J)Related Questions
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