Question

A student carried heated a 25.00 g piece of aluminum to a temperature of 100°C, and placed it in 100.00 g of water, initially at a temperature of 10.0°C. Determine the final temperature of the system (aluminum and water). Useful data: The specific heat of water is 4.18 J/(g°C) The specific heat of aluminum is 0.900 J/(g°C)

Answer 1

The final temperature of the system is 14.6 °C

Step-by-step explanation:

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Answer 2

The final temperature of the system is 14.6 °C

Step-by-step explanation:

Step 1: Data given

mass of the aluminium = 25.00 grams

mass of the water = 100.00 grams

Initial temperature of aluminium = 100 °C

Initial temperature of water = 10.0 °C

Specific heat of water = 4.18 J/g°C

Specific heat of aluminium 0.900 J/g°C

Step 2: Heat transfer

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(aluminium) * C(aluminium) * ΔT(aluminium) = - m(water) * C(water) * ΔT(water)

25*0.900*(T2-100) = - 100*4.18 * ( T2 - 10)

2250 - 22.5T2 = 418T2 - 4180

T2 =14.6 °C

The final temperature of the system is 14.6 °C