a metal sample having a temperature of 100 degrees Celsius and a mass of 45.2g was placed in 50g of water with a temperature of 26 degrees Celsius. at equilibrium, the final temperature was 32.7 degrees Celsius. a)how much heat flowed into the water?
b)what is the heat capacity of the metal?
To answer these questions, we can use the principle of heat transfer, specifically, the equation for heat transfer:
Q = mcΔT
Where:
Q is the heat transferred (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
Let's solve each part of the question step by step:
a) How much heat flowed into the water?
To calculate the heat flow into the water, we need to use the equation mentioned above.
The mass of water is given as 50g, and the change in temperature (ΔT) is calculated as the final temperature (32.7°C) minus the initial temperature (26°C):
ΔT = 32.7°C - 26°C = 6.7°C
Now, we can substitute the values into the equation:
Q = mcΔT
= (50g) * (4.18 J/g°C) * (6.7°C) [specific heat capacity of water is 4.18 J/g°C]
Calculating this, we get:
Q = 1,393 Joules
Therefore, the amount of heat flowed into the water is 1,393 Joules.
b) What is the heat capacity of the metal?
To calculate the heat capacity of the metal, we can use the equation:
Q = mcΔT
Here, the mass of the metal is given as 45.2g, and the change in temperature is the same as before, ΔT = 6.7°C.
Now, we need to rearrange the equation to solve for c (the specific heat capacity of the metal):
c = Q / (m * ΔT)
Substituting the values:
c = (1,393 Joules) / (45.2g * 6.7°C)
Calculating this, we get:
c ≈ 4.09 J/g°C
Therefore, the heat capacity of the metal is approximately 4.09 J/g°C.