The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114.3 J of heat. What is the
specific heat of iron

Q  =  mc(θ₂-θ₁)

47 calories =  10 g *c*(50.4 - 25)

47cal = 10*c* 25.4

47 /(10*25.4) = c

0.185 = c

Specific heat of iron = 0.185 cal/g°C

Explanation:

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explanation:

wind is defined by the direction it blows from.   for example, a wind blowing from north to south is called a northerly wind.

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