A 95 gram piece of metal is heated to 98°C and placed into 200mL of water insulated at 20°C. If the equilibrium temperature is 27.20, find:
The heat gained by the water. calories
The specific heat of the metal. cal/(g°C)
Hint: 1 mL water has a mass of 1 gram
I have no idea how to begin any of this, much less how to determine the answers for the two parts. Can someone help please?
heat gained by water:
masswater*specificheat*deltatemp
200g*4.18j/g*(27.2-20)=...
heat gained by water+heat change by metal=0
above heat calculation+95g*cmetal*(27.2-98)=0
solve for cmetal
6,019.2 calories for the heat gained by water is incorrect. That has been my problem all along. Can't figure out the correct answer.
Thanks, though.
To solve this problem, we need to apply the principle of heat transfer and use the formulas for heat gained or lost by a substance.
Let's break down the problem into smaller steps:
Step 1: Determine the heat gained by the water.
We can use the formula: Q = m * c * ΔT
where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the heat gained by the water can be calculated as follows:
m = 200 mL = 200 g (since 1 mL water has a mass of 1 gram)
c = specific heat capacity of water = 1 cal/(g°C) (given)
ΔT = equilibrium temperature - initial temperature = 27.20°C - 20°C = 7.20°C
So, the heat gained by the water is:
Q = (200 g) * (1 cal/(g°C)) * (7.20°C)
Q = 1440 cal
Therefore, the heat gained by the water is 1440 calories.
Step 2: Determine the specific heat of the metal.
To find the specific heat of the metal, we can use the formula:
Q = m * c * ΔT
In this case, the heat gained by the metal is equal to the heat lost by the water. So,
Q (metal) = -Q (water) (Since they have opposite signs)
Q (metal) = -1440 cal (Note the negative sign indicates a loss)
We are given the mass of the metal as 95 grams. So, substituting the values:
-1440 cal = (95 g) * c * (27.20°C - 98°C)
Now, we can solve for c:
c = -1440 cal / [(95 g) * (27.20°C - 98°C)]
c ≈ 0.1655 cal/(g°C)
Therefore, the specific heat of the metal is approximately 0.1655 cal/(g°C).
In summary:
1. The heat gained by the water is 1440 calories.
2. The specific heat of the metal is approximately 0.1655 cal/(g°C).