A lead calorimeter with mass of 150g contains 200g of water. The calorimeter and water are in thermal equilibrium at a specific temperature. Two metallic blocks are placed into the water. One is a 50.0g piece of aluminum heated at 90 degrees celcius. The other has a mass of 65g and is originally at a temperature of 97 degrees celcius. The entire system reaches a final temperature of 25 degrees celcius. (a) Find the final temp of the calorimeter and water. (b) Determine the specific heat of the sample.
To answer part (a) of the question, we need to use the principle of conservation of energy. The heat gained by the water, calorimeter, and the two metallic blocks should equal the heat lost by the aluminum block and the other metallic block.
The general formula for heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat transferred,
m is the mass of the substance,
c is the specific heat of the substance,
ΔT is the change in temperature.
Let's start by calculating the heat gained by the water and calorimeter.
1. Heat gained by water: Q_w = m_w * c_w * ΔT_w
Given: m_w = 200g (mass of water)
c_w = specific heat of water = 4.184 J/g°C (approximately)
ΔT_w = final temperature - initial temperature = 25°C - initial temperature
2. Heat gained by the calorimeter: Q_c = m_c * c_c * ΔT_c
Given: m_c = 150g (mass of calorimeter)
c_c = specific heat of calorimeter (assumed to be aluminum) = 0.897 J/g°C (approximately)
ΔT_c = final temperature - initial temperature = 25°C - initial temperature
Now, let's calculate the heat lost by the two metallic blocks.
3. Heat lost by the aluminum block: Q_al = m_al * c_al * ΔT_al
Given: m_al = 50.0g (mass of aluminum block)
c_al = specific heat of aluminum = 0.897 J/g°C (approximately)
ΔT_al = initial temperature - final temperature = 90°C - 25°C
4. Heat lost by the other metallic block: Q_m = m_m * c_m * ΔT_m
Given: m_m = 65g (mass of the other metallic block)
c_m = specific heat of the other metallic block (unknown)
ΔT_m = initial temperature - final temperature = 97°C - 25°C
Since the heat gained and lost should be equal, we can set up the equation:
Q_w + Q_c = Q_al + Q_m
(m_w * c_w * ΔT_w) + (m_c * c_c * ΔT_c) = (m_al * c_al * ΔT_al) + (m_m * c_m * ΔT_m)
Now, substitute the given values (except for c_m) and solve for ΔT_c.
Once we find ΔT_c, we can calculate the final temperature of the water and calorimeter:
final temperature = initial temperature + ΔT_c
To answer part (b) of the question, we can use the formula for specific heat:
c_m = Q_m / (m_m * ΔT_m)
Substitute the values of Q_m, m_m, and ΔT_m to calculate the specific heat of the other metallic block.