Question

A 3.6 kg piece of aluminum metal(c=900 J/kg°C) at 120°C is placed in 7.00 liters of water. If the system reaches a final temperature of 30°C, what was the heat transferred by the aluminum?

Answer 1

The heat transferred by the aluminum metal to the water can be calculated using the formula Q = mcΔT, resulting in a heat transfer of -291600 Joules, indicating that the aluminum lost this amount of heat.

Step-by-step explanation:

To calculate the heat transferred by the aluminum metal when it is placed in water, we can use the formula Q = mcΔT, where 'Q' is the heat transferred, 'm' is the mass, 'c' is the specific heat capacity, and ΔT is the change in temperature.

Given that the mass (m) of the aluminum is 3.6 kg, the specific heat capacity (c) of aluminum is 900 J/kg°C, and the temperature change (ΔT) is from 120°C to 30°C, we can calculate the heat transferred (Q) as follows:

Q = m • c • ΔT
Q = 3.6 kg • 900 J/kg°C • (30°C - 120°C)
Q = 3.6 kg • 900 J/kg°C • (-90°C)
Q = -291600 J

Since heat loss is considered as negative in thermal calculations, the aluminum transfers 291600 Joules of heat to the water.