19) A 100-g aluminum calorimeter contains 250g of water. The two substances are in thermal equilibrium at 10 C. Two metallic blocks are placed in the water. One is a 50-g piece of copper at 80 C. The other sample has a mass of 70g and is originally at a temperature of 100 C. The entire system stabilizes at a final temperature of 20 C. Determine the specific heat of the unknown second sample.
What is your question on this problem?
Heat is transferred from the two added blocks to the water and calorimeter. You can calculate that amount of heat from the increase in temperature. Set it equal to the heat lost by the two added blocks, and solve for the unknown specific heat of the second sample.
so it's just Heat(cold)=Heat(hot)?
Yes, if what you mean is
(heat gained by cold material) = (heat lost by hot material.)
Yes that is what I meant. Thank you, I figured it out.
94.08J/kg-1oC
To determine the specific heat of the unknown second sample, we can use the principle of conservation of energy. The heat lost by the hot object will be equal to the heat gained by the cold objects.
Let's calculate the heat lost by the hot object (Q_hot):
Q_hot = m_hot * c_hot * ΔT_hot
where:
m_hot is the mass of the hot object (70g)
c_hot is the specific heat of the hot object (unknown)
ΔT_hot is the change in temperature of the hot object (100°C - 20°C)
Now, let's calculate the heat gained by the cold objects (water and calorimeter):
Q_cold = (m_water + m_calorimeter) * c_water * ΔT_cold
where:
m_water is the mass of water (250g)
m_calorimeter is the mass of the calorimeter (100g)
c_water is the specific heat of water (4.186 J/g°C)
ΔT_cold is the change in temperature of the cold objects (20°C - 10°C)
Since the system is in thermal equilibrium, the heat lost by the hot object is equal to the heat gained by the cold objects, so we have:
Q_hot = Q_cold
m_hot * c_hot * ΔT_hot = (m_water + m_calorimeter) * c_water * ΔT_cold
Substituting the known values, we get:
70g * c_hot * (100°C - 20°C) = (250g + 100g) * 4.186 J/g°C * (20°C - 10°C)
Simplifying the equation:
70g * c_hot * 80°C = 350g * 4.186 J/g°C * 10°C
Dividing both sides by 80°C:
70g * c_hot = 350g * 4.186 J/g°C
Dividing both sides by 70g:
c_hot = (350g * 4.186 J/g°C) / 70g
Calculating the specific heat of the hot object:
c_hot = 20.09 J/°C