Question

a. An ice cold piece of aluminum metal is added to 50.0 g of hot water....

a. An ice cold piece of aluminum metal is added to 50.0 g of hot water. Given the average initial temperature (76 C) calculated above for the hot water, calculate the heat, q, in joules of the piece of aluminum metal if the final temperature of the water is 40.0 °C. The specific heat of water is 4.184 J/g-°C. (0.50) b. Calculate the grams of aluminum metal used if the specific heat of aluminum is 0.895 J/g-°C. (0.50)

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The aluminum cup inside your calorimeter weighs 39.96 g. You add 49.96 g of ice cold...
The aluminum cup inside your calorimeter weighs 39.96 g. You add 49.96 g of ice cold water to the calorimeter. You measure the temperature of the calorimeter to be 0.5oC just before your next addition. You then add 50.44 g of hot water and a 50.10 g metal object, all having an initial temperature of 69.5oC. After the calorimeter reaches thermal equilibrium, the final temperature is measured to be 36.1oC. Assume that: the calorimeter is completely insulated the heat capacity...
Calculate the specific heat of a metal from the following experimental data. 75.0 ml cold water...
Calculate the specific heat of a metal from the following experimental data. 75.0 ml cold water is taken in a calorimeter. The initial temp of the water in the calorimeter is 21.2 degrees C. To the calorimeter containing cold water 29.458 g metal at 98.9 degrees C is added. The final temperature of the contents of the calorimeter is measured to be 29.5 degreesC. (Given: density of water= 1.00 g/ml, specific heat of water= 4.184 J. G. -1 degrees C...
1. A 78.0 g piece of metal at 89.0°C is placed in 125 g of water...
1. A 78.0 g piece of metal at 89.0°C is placed in 125 g of water at 21.0°C contained in a calorimeter. The metal and water come to the same temperature at 27.0°C. - How much heat (in J) did the metal give up to the water? (Assume the specific heat of water is 4.18 J/g·°C across the temperature range.) - What is the specific heat (in J/g·°C) of the metal? 2. A 0.529 g sample of KCl is added...
Suppose that 100.0 g of ice at 0 degrees Celsius are added to 300.0 g of...
Suppose that 100.0 g of ice at 0 degrees Celsius are added to 300.0 g of water at 25.00 degrees Celsius. Is this sufficient ice to lower the temperature of the water to 5.00 degrees Celsius and still have ice remaining? Calculate the energy (heat), which must be removed from water to achieve the desired temperature change, and then prove that there is (is not) sufficient ice to cool the water. Use the specific heat capacity of water (4.184 J/g-*C)...
An 100-g aluminum calorimeter contains 280 g of water at an equilibrium temperature of 20°C. A...
An 100-g aluminum calorimeter contains 280 g of water at an equilibrium temperature of 20°C. A 170-g piece of metal, initially at 277°C, is added to the calorimeter. The final temperature at equilibrium is 32°C. Assume there is no external heat exchange. The specific heats of aluminum and water are 910 J/kg·K and 4190 J/kg·K, respectively. The specific heat of the metal is closest to: a) 270 J/kg·K. b) 240 J/kg·K. c) 330 J/kg·K. d) 390 J/kg·K. e) 360 J/kg·K.
A hot lump of 26.3 g of aluminum at an initial temperature of 67.2 °C is...
A hot lump of 26.3 g of aluminum at an initial temperature of 67.2 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
803 cal of heat is added to 5.00 g ice at –20.0 °C. What is the...
803 cal of heat is added to 5.00 g ice at –20.0 °C. What is the final temperature of the water? SPecific heat H2O(s)= 2.087 J/(g*C) Specific heat H2O(l)=4.184 J/(g*C) Heat of fusion= 333.6 J/g
A hot lump of 42.6 g of aluminum at an initial temperature of 62.2 °C is...
A hot lump of 42.6 g of aluminum at an initial temperature of 62.2 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings. Please show all work
A piece of titanium metal with a mass of 20.8 g is heated in boiling water...
A piece of titanium metal with a mass of 20.8 g is heated in boiling water to 99.5 0C and then dropped into a coffee cup calorimeter containing 75.0 g of water at 21.7 0C.When thermal equilibum is reached, the final temperature is 14.30C.Calculate the specific heat capacity of titanium. ( Specific Heat Capacity of H2O (l) =4.184 J g-1 0C-1)
A metal sample weighing 72.1 g is placed in a hot water bath at 95.0 oC....
A metal sample weighing 72.1 g is placed in a hot water bath at 95.0 oC. The calorimeter contains 42.3 g of deoinized water. The initial temperature of the water is 22.3 oC. The metal is transferred to the calorimeter and the final temperature reached by the water + metal is 32.2 oC. A. Calculate ∆T for the water (Tfinal – Tinitial). B. Calculate ∆T for the metal. C. The specific heat of water is 4.18 J/goC. Calculate the specific...