Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
The mass of the steel is 41.9 grams
To calculate the mass of the steel, we make use of the law of conservation of energy.
Mathematically, it is represented as:
[tex]Q_a + Q_b = 0[/tex]
Where Q represents the amount of heat.
Let w represents water (H2O) and s represents steel.
The equation becomes
[tex]Qw + Q_s = 0[/tex]
The amount of heat is calculated using:
[tex]Q =mc\Delta T[/tex]
Where:
c = the specific heat capacity.
For water, we have:
[tex]Q_w =m_w * c_w * \Delta T_w[/tex]
[tex]Q_w =m_w * 4.18 * (21.30 - 22.00)[/tex]
[tex]Q_w =-m_w * 2.926[/tex]
The mass of water is calculated as:
[tex]m_w = density * volume[/tex]
The density of water is approximately 1.00 g/mL
So, we have:
[tex]m_w = 1.00gmL^{-1} * 125mL[/tex]
This gives
[tex]m_w = 125g[/tex]
The equation [tex]Q_w =-m_w * 2.926[/tex] becomes
[tex]Q_w =-125 * 2.926[/tex]
[tex]Q_w =-365.75[/tex]
For steel, we have:
[tex]Q_s =m_s * c_s * \Delta T_s[/tex]
[tex]Q_s = 0.452 * m_s * (21.30-2.00)[/tex]
This gives
[tex]Q_s = 8.7236m_s[/tex]
Recall that:
[tex]Qw + Q_s = 0[/tex]
So, we have:
[tex]-365.75 + 8.7236m_s = 0[/tex]
Collect like terms
[tex]8.7236m_s = 365.75[/tex]
Solve for ms
[tex]m_s = 41.9[/tex] -- approximated
Hence, the mass of the steel is 41.9 grams
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