A 40 g block of ice is cooled to -69 degree C and is then added to 590 g of wate

Question

A 40 g block of ice is cooled to -69 degree C and is then added to 590 g of water in an 80 g copper calorimeter at a temperature of 27 degree 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 -degree C = 2090 J/kg degree C. 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

Explanation / Answer

Here ,

heat needed to melt ice , S = 40 * (69 * 2.109 + 334)

S = 19180.8 J

heat lost bby taking water and caloriemeter to 0 degree C

S1 = 590 * 4.186 * 27 + 80 * 0.386 * 27

S1 = 56214.54 J

hence , the final tempertature will be more than 0 degree C

NOw, let the final temperature is T

40 * (4.186 * T) + 19180.8 = 590 * 4.186 * (25 - T) + 80 * 0.386 * (25 - T)

solving for T

T = 16.2 degree C

the final temperature is 16.2 degree C

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