Two bars of identical mass are at 30 °C. One is made from glass and the other from another substance. The specific heat capacity of glass is 840 J/(kg · C°). When identical amounts of heat are supplied to each, the glass bar reaches a temperature of 95 °C, while the other bar reaches 292.0 °C. What is the specific heat capacity of the other substance?
Q₁=mc₁ΔT₁
Q₂=mc₂ΔT₂
mc₁ΔT₁=mc₂ΔT₂
c₁ΔT₁=c₂ΔT₂
c₁(95-30)=c₂(292-30)
c₂=c₁(95-30)/(292-30)=
=840•65/262 = 208.4 J/kg•°C.
Well, well, well, looks like we have two hot bars on our hands! One made of glass, and the other made of... well, who knows? Let's figure it out, shall we?
First, we need to find the temperature change for both bars. For the glass bar, it goes from 30 °C to 95 °C, so the temperature change is 95 °C - 30 °C = 65 °C. As for the other bar, it goes from 30 °C to 292 °C, so the temperature change is 292 °C - 30 °C = 262 °C.
Now, let's apply some heat (you better not melt the bars, okay?). We have identical amounts of heat supplied to both bars, but they reach different temperatures. So, we need to consider their specific heat capacities.
For the glass bar, the specific heat capacity is given as 840 J/(kg·C°). Since we don't know the mass of the bars, we can call the mass m and assume it cancels out when comparing the two bars.
Using the formula Q = mcΔT (Q is the heat supplied, m is the mass, c is the specific heat capacity, and ΔT is the temperature change), we have:
Q for the glass bar = mcΔT
Q for the other bar = mc'ΔT
Since the heat supplied to both bars is identical, we have:
mcΔT = mc'ΔT
The masses cancel out, leaving us with:
cΔT = c'ΔT
Simplifying further, we have:
c = c'
So, the specific heat capacity of the other substance is the same as that of glass, specifically 840 J/(kg·C°).
And there you have it! Now you know the hot gossip on the specific heat capacity of the other bar. It's just as cool as glass.
To solve this problem, we can use the formula for heat transfer:
q = mcΔT
where q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Let's assume both bars have the same mass, so we can cancel out the mass term from the equation.
Now, we can focus on the two cases separately:
1. For the glass bar:
q1 = c1 * ΔT1
where c1 is the specific heat capacity of glass, and ΔT1 is the change in temperature of the glass bar.
Given that ΔT1 = 95 °C - 30 °C = 65 °C, and c1 = 840 J/(kg · °C), we can calculate q1.
2. For the other substance bar:
q2 = c2 * ΔT2
where c2 is the specific heat capacity of the other substance, and ΔT2 is the change in temperature of the other substance bar.
Given that ΔT2 = 292.0 °C - 30 °C = 262.0 °C, we can calculate q2.
Since both bars are supplied with identical amounts of heat, we can assume that q1 = q2.
Now, we can equate the two equations:
c1 * ΔT1 = c2 * ΔT2
Solving for c2, the specific heat capacity of the other substance:
c2 = (c1 * ΔT1) / ΔT2
Substituting the given values, we have:
c2 = (840 J/(kg · °C) * 65 °C) / 262.0 °C
Calculating this equation will give us the specific heat capacity (c2) of the other substance.
To find the specific heat capacity of the other substance, we need to use the formula:
Q = mcΔT
where Q is the heat energy supplied to the substance, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Since the masses of the bars are identical, we can set up the equation:
mcΔT for glass = mcΔT for other substance
The initial temperature of both bars is 30 °C, so their change in temperature are:
ΔT for glass = 95 °C - 30 °C = 65 °C
ΔT for other substance = 292 °C - 30 °C = 262 °C
We know that the specific heat capacity of glass is 840 J/(kg · C°). Let's assume the specific heat capacity of the other substance is c2. The equation becomes:
m * 840 * 65 = m * c2 * 262
Since the masses of the bars are identical, m cancels out:
840 * 65 = c2 * 262
Now we can solve for c2:
c2 = (840 * 65) / 262
Calculating this gives us:
c2 = 210 J/(kg · C°)
Therefore, the specific heat capacity of the other substance is 210 J/(kg · C°).