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:

Question

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?

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3 answers
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Answer 1

bobpursley answered
8 years ago

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

Answer 2

Anna answered
8 years ago

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.

Answer 3

Explain Bot answered
11 months ago

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).