Question

a substance is heated from 15°C to 18°C by providing heat of 270 calories. If mass of the substance is 3 kg , determine specific heat of the substance.?




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Answer 1

Given :

• Initial temperature = 15°C
• Final temperature = 18°C
• Heat supplied = 270 calories
• Mass of substance = 3kg

To find :

Specific heat of substance

Formula used :

Q = mc∆T

Here,

• Q = Heat supplied
• m = Mass of substance
• c = Specific heat of substance
• ∆T = Temperature change

Solution :

Firstly we need to find temperature change

➝ Temperature change = Final temperature - Initial temperature

➩ ∆T = 18°C - 15°C

➩ ∆T = 3°C

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➝ Mass = 3kg

➩ Mass = 3kg × 1000g/kg

➩ Mass = 3000 g

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Now putting value in above formula, we get,

➝ 270 cal = 3000g × c × 3°C

➩ 270 cal = 9000(g°C) × c

➩ c = 270 cal ÷ [ 9000 g°C]

➩ c = (270/9000) cal/(g°C)

➩ c = 0.03 cal/ (g°C)

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ANSWER :

Specific heat of substance = 0.03 cal/(g°C)

Answer 2

Explanation:

Heat of substance = 0.03 cal/(g°c)

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