A small 36 g water sample A is heated for 234 seconds by a 19.3 Watt heater resu

Question

A small 36 g water sample A is heated for 234 seconds by a 19.3 Watt heater resulting in an increase temperature capital delta T subscript A . The heated small water sample A is added to a larger tank of water, sample B, but, during the pouring process, water sample A cools down by 10% (a 10% decrease in the capital delta T subscript A ). The larger tank of water sample B was originally 394 g of water, before sample A was added. What is the final temperature change capital delta T subscript A plus B end subscript of the combined samples? Units are in degrees Celsius; do not include units in your answer. Use Cp = 4.1813 J/goC, and give your answer to two decimal places

Explanation / Answer

Mass of the water sample A (m)= 36 g

Mass of the water sample B (M)= 394 g

Power of the heater P = 19.3 watt = 19.3 J/s

Time for which sample A is heated t= 234 s

So heat supplied to sample A (Q) = P * t

                      Q = 19.3 * 234 = 4516.2 J

Let TA be rise in the temperature

Then TA = Q/Cp * m = 4516.2/36 * 4.1813

                                  = 300 C

We are given that there is a fall of 10% in the temperature after it is mixed with sample B

Fall in temperature = 30 * 10/100 = 30 C

Final temperature of sample A is 300 – 30 = 270 C

So final and common temperature of the sample is 270 C

Now the common temperature is given by

( TA + TB ) = ( ( m * TA )+( M * TB ))/ ( m + M )

Where TB is temperature of sample B

27 = ( 36 * 30 )+ ( 394 * TB)/( 36 + 394 )

27 * 430 = 1080 + ( 394 * TB )

394 * TB = 11610 – 1080

TB = 10530/394 = 26.70 C

So rise l in temperature of sample B = 270 - 26.70 = 0.30 C

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