equal masses m of stone and wood reach thermal equilibrium at T=290. the initial temperature of the wood reach is T=285. find the initial temperature of the stone.(specific heat of wood=1764 and specif heat of stone =882)
To find the initial temperature of the stone, we can use the principle of thermal equilibrium, which states that when two objects are in contact, heat will flow from the object with higher temperature to the one with lower temperature until their temperatures become equal.
Let's denote the initial temperature of the stone as Ts. We know that the final temperature of both the stone and wood is 290 degrees.
We can use the equation for heat transfer:
Q = mcΔT
Where Q is the heat transfer, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
For the stone:
Qstone = m * cstone * (290 - Ts)
For the wood:
Qwood = m * cwood * (290 - 285)
Since the two objects reach thermal equilibrium, the heat lost by the stone is equal to the heat gained by the wood, so:
Qstone = Qwood
m * cstone * (290 - Ts) = m * cwood * (290 - 285)
We can cancel out the mass on both sides of the equation:
cstone * (290 - Ts) = cwood * (290 - 285)
Plugging in the specific heat capacities:
882 * (290 - Ts) = 1764 * (290 - 285)
Now, let's solve for Ts.
882 * 290 - 882 * Ts = 1764 * 290 - 1764 * 285
882 * Ts = 1764 * (290 - 285)
Dividing both sides by 882:
Ts = 1764 * (290 - 285) / 882
Ts = 8820 / 882
Ts = 10
Therefore, the initial temperature of the stone, Ts, is 10 degrees Celsius.