Question

A 126.1-gram block of granite at 92.6°c is dropped into a tub of water at 24.7°c in an isolated system. the final temperature of both the granite and the water is 51.9°c. the specific heat capacity of granite is 0.795 joules/gram degree celsius, and the specific heat capacity of water is 4.186 joules/gram degree celsius. the granite block transferred of energy, and the mass of the water is .

Answer 1

The granite block transferred 4,080 joules of energy, and the mass of the water is 35.8 grams.

How we calculate heat transfer of any system ?

Heat transfer in any system can be calculated as follow:

q = mCΔT

Given ;

• m = mass of granite = 126.1 g
• C = specific heat of granite = 0.795 joules/gram degree Celsius

• ΔT = change in temperature of granite = 51.9 °C - 92.6 °C = -40.7 °C

Putting all the values in folowing equation,

q = mCΔT

we get

q = 126.1 × 0.795 × -40.7 = -4080 J

Now by using same formula, we can also calculate the mass of water in the following way:

m = q / CΔT

where

• C = specific heat of water = 4.186 joules/gram degree Celsius
• ΔT = change in temperature of water = 51.9 °C – 24.7 °C = 27.2 °C

Putting all these values ;

m = -4080 /  4.186 × 27.2 = 35.8 g.

Hence, 4,080 joules of energy is transferred by granite block and 35.8 g is the mass of water.

To learn more about heat transfer here ;

brainly.com/question/18980657

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