Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 C, as the heat provided when 100g of steam is cooled rom 110C to 100C.

Answer:

8.08 grams

Explanation:

Given;

Mass of the steam, m = 100 g

Change in temperature of water = (95°C - 35°C) = 60°C

Change in temperature of steam = (110°C - 100°C) = 10°C

let the mass of water by 'M'

also,

We know

specific heat of water, c = 4.184 J/Kg.°C

And, specific heat of steam, C = 2.03 J/Kg.°C

Now,

Heat released = Mass × specific heat × change in temperature

thus, for the condition given in the question

M × 4.184 × 60 = 100 × 2.03 × 10

or

M = 8.08 grams

Final answer:

The mass of water that will provide the same amount of heat when cooled from 95.0 to 35.0 C as a 100 g of steam cooling from 110C to 100C is 125 grams. To solve this question, we used the specific heat of water, calculated the heat produced in the latter scenario, and deduced the mass needed from that.

Explanation:

To determine the mass of water that will provide the same amount of heat when cooled from 95.0 to 35.0 C as a 100 g of steam cooling from 110 to 100C, we need to first calculate the amount of heat produced in the latter situation. Using the specific heat of water which is 4.184 J/g °C, the heat produced when cooling 100g of steam from 110C to 100C is Q=msΔT, where m is mass, s is specific heat, and ΔT is temperature difference. So, Q = 100g * 4.184 J/g°C * (110°C - 100°C) = 4184 J.

Now, we have to calculate the mass of water that will produce the same amount of heat when cooled from 95°C to 35°C. We rearrange the equation to solve for mass. Hence, m = Q/sΔT = 4184 J / 4.184 J/g°C / (95.0 °C - 35.0 °C) = 125g. Therefore, 125g of water will provide the same amount of heat when cooled from 95.0 to 35.0 C.

Learn more about Heat Transfer here:

brainly.com/question/13433948

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