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A sample of metal with a specific heat capacity of 0.50 kk/kg.•c Is heated to 98 degrees c and then placed in an 0.055 kg sample of water at 22 degrees C

2 Answers

4 votes

Answer:

We must identify the heat transfers that are happening here.

One is the heat transferred from the metal as it cools (

q

1

).

The second is the heat transferred to the water as it warms (

q

2

).

Per the Law of Conservation of Energy, the sum of the two heat transfers must be zero.

Step-by-step explanation:

User ChyBy
by
3.9k points
9 votes

The final temperature of the metal and water is 75.6 ⁰C.

How to calculate the final temperature of the mixture?

The final temperature of the metal and water is calculated by applying the principle of conservation of energy.

Heat obtained by the water equals heat lost by the metal.

The following is the formula for heat capacity:

Q = mcΔθ

where;

  • m is the mass of the metal
  • c is the specific heat capacity
  • Δθ is the change in temperature

The final temperature of the metal and water is calculated as;

1 kg x 500 J/kgC x (98 - T) = 0.05 kg x 4186 J/kgC x (T - 22)

49,000 - 500T = 209.3T - 4,604.6

49,000 + 4604.6 = 209.3T + 500T

709.3 T = 53,604.6

T = 53,604.6 / 709.3

T = 75.6 ⁰C

The complete question is below:

A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 98 ⁰C and then placed in a 50.0 g sample of water at 22 ⁰C. What is the final temperature of the metal and the water?

User Sachin Malhotra
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