1. A 0.400 kg sample of aluminum (c = 9.20 x 102 J/kg°C) at 95.0 °C is dropped into a 0.550 kg pot of water which is at 18.0 °C. What temperature will the mixture come to?
2. How much thermal energy is absorbed by a 0.0220 kg ice cube as it melts? The latent heat of fusion of water is 3.40x 10^5 J/kg.
Equation: Qlost + Qgained = 0
m1 c1∆T1+m2 c2 ∆T2=0
Substitute:
(0.550kg)(4200 J/kg℃)(T2-18.0 ℃)+(0.400 kg)(920 J/kg℃)(T2-95.0 ℃)=0
2310(T2-18.0 ℃)+368(T2-95.0 ℃)=0
2310T2-41580+368T2-34960=0
2678T2-76540=0
2678T2=76540
T2=28℃
Statement: The mixture will come to an equilibrium temperature of 28℃.
2.
Given: m=0.0220 kg,Lf=3.40x10^5 J/kg
Unknown: Q=?
Equation: Q=mLf
Substitute: Q=(0.0220 kg)(3.40x10^5 J/kg)
Q=7480
Q=7480 J
Statement: Therefore, 7.5x10^3 J worth of thermal energy is absorbed by the ice as it melts.
Wazzzuuup
1. Well, here's a math problem that's a real hot topic! We have some aluminum and water, ready to mingle. So, the heat gained by the water (m1•c1•ΔT1) must equal the heat lost by the aluminum (m2•c2•ΔT2). Since the final temperature is the same for both, we can set them equal to each other and solve for that hot temperature!
(m1•c1•ΔT1) = (m2•c2•ΔT2)
(0.550 kg)(4186 J/kg°C)(T - 18.0 °C) = (0.400 kg)(920 J/kg°C)(T - 95.0 °C)
Now, I could give you the answer right away, but where's the fun in that? I'll leave the math to you – remember, the temperature will be in Celsius. Good luck with your thermal love affair!
2. Ah, the icy side of things. We have an ice cube that's going on a liquid adventure! To determine the amount of thermal energy absorbed, we can multiply the mass of the ice cube (m) by the latent heat of fusion (L).
Energy absorbed = mass × latent heat of fusion
Now, let's plug in the numbers:
Energy absorbed = (0.0220 kg)(3.40 x 10^5 J/kg)
And there you have it, the amount of thermal energy absorbed by our melting ice cube. Just be careful not to slip on any puns along the way!
1. To solve this problem, we can use the principle of conservation of energy. The heat lost by the aluminum sample will be equal to the heat gained by the water.
First, let's calculate the heat lost by the aluminum sample using the equation:
Q1 = m1 * c1 * ΔT1
Where:
Q1 is the heat lost by the aluminum sample,
m1 is the mass of the aluminum sample (0.400 kg),
c1 is the specific heat capacity of aluminum (9.20 x 10^2 J/kg°C),
and ΔT1 is the change in temperature of the aluminum sample (final temperature - initial temperature).
ΔT1 = final temperature - initial temperature = Tf - Ti = Tf - 95.0°C
Next, let's calculate the heat gained by the water using the equation:
Q2 = m2 * c2 * ΔT2
Where:
Q2 is the heat gained by the water,
m2 is the mass of the water (0.550 kg),
c2 is the specific heat capacity of water (4.18 x 10^3 J/kg°C), and
ΔT2 is the change in temperature of the water (final temperature - initial temperature).
ΔT2 = final temperature - initial temperature = Tf - 18.0°C
Since the heat lost by the aluminum sample is equal to the heat gained by the water, we can set Q1 equal to Q2:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Now let's substitute the known values and solve for the final temperature (Tf):
0.400 kg * (9.20 x 10^2 J/kg°C) * (Tf - 95.0°C) = 0.550 kg * (4.18 x 10^3 J/kg°C) * (Tf - 18.0°C)
Simplify this equation and solve for Tf:
0.400 * 9.20 x 10^2 * Tf - 0.400 * 9.20 x 10^2 * 95.0 = 0.550 * 4.18 x 10^3 * Tf - 0.550 * 4.18 x 10^3 * 18.0
Now rearrange this equation to solve for Tf:
0.400 * 9.20 x 10^2 * Tf - 0.550 * 4.18 x 10^3 * Tf = 0.550 * 4.18 x 10^3 * 18.0 - 0.400 * 9.20 x 10^2 * 95.0
Now solve for Tf.
2. To calculate the thermal energy absorbed by the ice cube as it melts, we can use the equation:
Q = m * ΔH
Where:
Q is the thermal energy absorbed,
m is the mass of the ice cube (0.0220 kg),
and ΔH is the latent heat of fusion of water (3.40 x 10^5 J/kg).
Now, plug in the known values and solve for Q:
Q = 0.0220 kg * (3.40 x 10^5 J/kg)
To find the temperature the mixture will come to in the first question, we need to use the principle of conservation of energy. The heat gained by the aluminum will be equal to the heat lost by the water.
Let's calculate the heat gained by the aluminum first.
Step 1: Calculate the heat gained by the aluminum:
Using the formula Q = mcΔT, where Q is the heat gained (or lost), m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature:
Q(aluminum) = (0.400 kg) * (9.20 × 10^2 J/kg°C) * (T - 95.0 °C), where T is the final temperature of the mixture.
Step 2: Calculate the heat lost by the water:
Using the same formula, Q(