Apparently, I\'m close, but I cannot figure out where my 10% error is... A 40 g

Question

Apparently, I'm close, but I cannot figure out where my 10% error is...

A 40 g block of ice is cooled to -76degree C and is then added to 610 g of water in an 80 g copper calorimeter at a temperature of 23degree C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0 degree C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g middot degree C = 2090 J/kg degree C. 14.82608 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. Degree C A Styrofoam box has a surface area of 0.83 m^2 and a wall thickness of 2.0 cm. The temperature of the inner surface is 5.0degree C, and the outside temperature is 26degree C. If it takes 8.6 h for 5.0 kg of ice to melt in the container, determine the thermal conductivity of the Styrofoam. 0.0617 W/m middot K A bar of lead (bar A) is in thermal contact with a bar of silver (bar B) of the same length and area. One end of the compound bar is maintained at T_h = 87.0degree C while the opposite end is at 30.0degree C. Find the temperature at the junction when the energy flow reaches a steady state. 34.3 degree C

Explanation / Answer

2090 J/(kg°C) = 0.5 cals/(g°C)
Heat lost by calorimeter to fall to° C= 80*0.5* (23-)
Heat lost by water to fall to ° C = 610*(23-)

40 g of needs 40*(0.5*76 +80) cals = 4720 to convert into ice at 0° C
Heat gained by ice = 40(0.5*76 +80 + 1*)

40*(0.5*76 +80 + ) = (80*0.5 + 610)* (23-)

= 14.83°C

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