High School

A 5.00 g sample of aluminum pellets (specific heat capacity = 0.89 J/°C·g) and a 10.00 g sample of iron pellets (specific heat capacity = 0.45 J/°C·g) are heated to 100.0 °C. The mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.0 °C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Answer :

Final answer:

The final temperature of the system consisting of the two metal samples and the water is 34.0°C.

Explanation:

Let's first calculate the heat gained by the water. Given that the mass of water is 50.0 g, the specific heat capacity of water is 4.18 J/g°C, the final temperature is unknown, and the initial temperature is 25.0 °C, we can substitute these values into the equation Q = m * c * (T`f - Ti). Rearranging the equation and solving for T`f:

Q = m * c * (T`f - Ti)

(50.0 g) * (4.18 J/g°C) * (T`f - 25.0 °C) = (10.00 g * 0.897 J/g°C * (T`f - 25.0 °C) + 30.0 g * 0.39 J/g°C * (T`f - (-20.0 °C)))

Now we can solve for T`f. Let's do the calculation:

T` f = 34.0 °C

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