Three 75.0-g samples of copper, silver, and gold are available. Each of these samples is intially at 24.0 degrees C, and 2.00 kJ of heat is applied to each sample. Which sample will end up at the highest temperature?
q = mass x specific heat x (Tfinal-Tintial)
2000 = 75 x specific heat x delta T
specific heat = 2000/(75*delta T)
high sp.h. = lowest delta T
low sp.h. = highest delta T
To determine which sample will end up at the highest temperature, we can calculate the final temperature using the formula:
q = mcΔT
where:
q = heat transferred to the sample (in joules)
m = mass of the sample (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Given:
Mass of the copper sample (m_copper) = 75.0 g
Mass of the silver sample (m_silver) = 75.0 g
Mass of the gold sample (m_gold) = 75.0 g
Heat applied to each sample (q) = 2.00 kJ = 2.00 * 10^3 J
First, let's calculate the change in temperature for each sample using the formula ΔT = q / (mc):
For copper:
ΔT_copper = q / (m_copper * c_copper)
For silver:
ΔT_silver = q / (m_silver * c_silver)
For gold:
ΔT_gold = q / (m_gold * c_gold)
Now, let's compare the ΔT values to find out which sample will end up at the highest temperature.
To determine which sample will end up at the highest temperature, we need to calculate the final temperature for each sample based on the amount of heat applied.
To solve this problem, we can use the heat transfer formula:
Q = mcΔT
Where:
Q = heat energy transferred
m = mass
c = specific heat capacity
ΔT = change in temperature
We need to find the final temperature (ΔT) for each sample, assuming that they all gain the same amount of heat (2.00 kJ). We can rearrange the formula to find the change in temperature:
ΔT = Q / (mc)
Now, let's calculate the final temperature for each sample:
For copper:
m = 75.0 g
copper specific heat capacity (c) = 0.385 J/g·°C
Q = 2.00 kJ = 2000 J
ΔT(copper) = 2000 J / (75.0 g * 0.385 J/g·°C)
ΔT(copper) ≈ 69 °C
For silver:
m = 75.0 g
silver specific heat capacity (c) = 0.235 J/g·°C
ΔT(silver) = 2000 J / (75.0 g * 0.235 J/g·°C)
ΔT(silver) ≈ 113 °C
For gold:
m = 75.0 g
gold specific heat capacity (c) = 0.129 J/g·°C
ΔT(gold) = 2000 J / (75.0 g * 0.129 J/g·°C)
ΔT(gold) ≈ 197 °C
Comparing the ΔT values for each sample, we can see that gold will end up at the highest temperature. Therefore, the gold sample will end up at the highest temperature.