A 2.00 kg piece of granite with a specific heat of 0.803 J/g·°C and a temperature of 95.0°C is placed
into 2.00 L of water at 22.0°C. When the granite and water come to the same temperature, what
will that temperature be?
To find the final temperature when the granite and water come to equilibrium, we need to use the concept of heat transfer.
The heat gained or lost by an object can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat gained or lost (in Joules)
m is the mass of the object (in grams)
c is the specific heat of the object (in J/g·°C)
ΔT is the change in temperature (in °C)
Now, let's calculate the heat gained or lost by the granite and the water.
For the granite:
Mass (m) = 2.00 kg × 1000 g/kg = 2000 g
Specific heat (c) = 0.803 J/g·°C
Initial temperature (T1) = 95.0°C
Final temperature (Tf) = ?
For the water:
Volume (V) = 2.00 L = 2000 mL = 2000 g (since 1 mL = 1 g for water)
Specific heat (c) = 4.18 J/g·°C (specific heat of water)
Initial temperature (T1) = 22.0°C
Final temperature (Tf) = ?
Now, let's calculate the heat gained or lost by the granite (Qg) and the water (Qw):
Qg = mcΔTg
Qw = mcΔTw
The heat lost by the granite will be equal to the heat gained by the water. Therefore:
Qg = -Qw
mcΔTg = -mcΔTw
Since the masses (m) and specific heats (c) are the same, we can cancel them out:
ΔTg = -ΔTw
We know that the initial temperature of the granite is 95.0°C and the initial temperature of the water is 22.0°C. Let's substitute those values:
ΔTg = - (Tf - 22.0°C)
95.0°C - Tf = - (Tf - 22.0°C)
Simplifying the equation:
95.0°C - Tf = -Tf + 22.0°C
95.0°C - 22.0°C = -Tf + Tf
73.0°C = Tf
Therefore, the final temperature when the granite and water come to the same temperature will be 73.0°C.