Heat absorbed(q) = ms(∆T)
q = 95.9×0.930×(67-23)
= 3924.2 J
Heat will be absorbed = 3924.2 J
Question 11 of 20 Un How much heat will be absorbed by a 95.9 g piece...
How much heat will be absorbed by a 72.3 g piece of aluminum (specific heat = 0.930 J/g・°C) as it changes temperature from 23.0°C to 67.0°C?
3. Calculate the heat absorbed by 25.0 g of water needed to raise its temperature from 20.0°C to 65.0°C. The specific heat of water is 4.18 J/gºC. Show your work Final Answer 4. Aluminum metal has a specific heat of 0.901 J/g C. How much heat is transferred to a 6.75 g piece of aluminum initially at room temperature, 20.0°C, when it is placed into boiling water? The temperature of boiling water is 100°C. Show your work Final Answer 5....
How much heat (in J) is evolved or absorbed when a 35.3 g sample of aluminum (specific heat = 0.900 J/(g K)) cools from 85.9oC to 69.5oC?
3. Calculate the heat absorbed by 25.0 g of water needed to raise its temperature from 20.0°C to 65.0'C. The specific heat of water is 4.18 J/gºC. Show your work Final Answer 4. Aluminum metal has a specific heat of 0.901 J/g C. How much heat is transferred to a 6.75 g piece of aluminum initially at room temperature, 20.0°C, when it is placed into boiling water? The temperature of boiling water is 100°C. Show your work Final Answer 5....
4) If 46.2 g piece of aluminum is cooled from 84.5 °C to 29.5C, how much energy was lost by aluminum? (specific heat of aluminum is 0.901 J/(g.°C)) 5) If 78.6 g piece of iron is dropped into 100.0 g of water initially at 24.6 °C. The final temperature of both water and iron was measure at 28.5 °C. Calculate the initial temperature of iron? (Specific heat of water and iron from #1)
please write clearly and use significant figures. thanks in
advance
3. Calculate the heat absorbed by 25.0 g of water needed to raise its temperature from 20.0°C to 65.0°C. The specific heat of water is 4.18 J/gºC. Show your work Final Answer 4. Aluminum metal has a specific heat of 0.901 J/gºC. How much heat is transferred to a 6.75 g piece of aluminum initially at room temperature, 20.0°C, when it is placed into boiling water? The temperature of boiling...
Question 10 of 20 A 141.9 g piece of copper (specific heat 0.38 J/g.°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g °C).
The specific heat of aluminum is 0.900 J/g*C. How many joules of heat are absorbed by 15.0 grams of AI if it is heated from 20.0 C to 60.0 C. Please show how to work out.
A) How much heat must be absorbed by a 23.0 g sample of water to raise its temperature from 30.0 ∘C to 50.0 ∘C? (For water, Cs=4.18J/g∘C.) Express your answer to three significant figures and include the appropriate units. ΔErxn = nothing kJ/mol B) anganese reacts with hydrochloric acid to produce manganese(II) chloride and hydrogen gas. Mn(s)+2HCl(aq) → MnCl2(aq)+H2(g) You may want to reference (Page 271) Section 6.7 while completing this problem. Part A When 0.635 g Mn is combined...
3) If 28.0 g of water at 26.4 °C gain 5,563 J of heat, what is the final temperature of water? (Specific heat of water is 4.184 J/g.°C)) 4) If 46.2 g piece of aluminum is cooled from 84.5 °C to 29.5 °C, how much energy was lost by aluminum? (specific heat of aluminum is 0.901 J/(g.°C)) 5) If 78.6 g piece of iron is dropped into 100.0 g of water initially at 24.6 °C. The final temperature of both...