Respuesta :

Answer: 80.6

Explanation:

q = mcAt

we have the grams of water, temp change, and specific heat of water is 4.186 J/g so we can plug in our numbers to find the heat energy in joules

q = (3.10)(4.184)(82.4-56.4)

q= 337.2304 J

we need our answer to be in calories so divide our answer by 4.184 and we get 80.6 calories

samueladesida43

The amount of calorie required to warm 3.10 g of water from 56.4°C to 82.4°C is 80.6 calories.

HOW TO CALCULATE CALORIES:

  • The amount of calories can be calculated by using the formula as follows:

Q = m × c × ∆T

Where:

  1. Q = amount of heat in calories
  2. c = specific heat capacity (J/g°C)
  3. ∆T = change in temperature (°C)
  4. m = mass (g)

  • According to this question, c = 4.184J/g°C, m = 3.10g, ∆T = 82.4°C - 56.4°C = 26°C

  • Q = 3.10 × 4.184 × 26

  • Q = 337.23J

  • Since 1J = 0.2390057361 calories

  • 337.23 J = 80.6calories.

  • Therefore, amount of calorie required to warm 3.10 g of water from 56.4°C to 82.4°C is 80.6 calories.

Learn more at: https://brainly.com/question/8097823?referrer=searchResults

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