Question

A sample of stainless steel was stored in a freezer with a
temperature of -4.0°C. After removing it from the freezer, the
aluminum is placed in 1982 mL of water at room temperature
(25.0°C). After several minutes thermal equilibrium is achieved at
21.4°C. What is the mass of the *stainless steel sample?*
Please show all work with appropriate units.

Answer #1

Let mass of stainless Steel is x grams

By using conservatio low of heat

Heat change in stainless steel + cheat change in water = 0

M(s)×C(s)×∆T(s) + M(w)×C(w)×∆T(w)

where, M(s) = mass of stainless steel = x g

C(s) = specific heat capacity of stainless steel = 0.502J/g-°C

∆T(s) = temperature change for stainless steel = 21.4 -(-4) = 25.4°C

M(w) = mass of water

(We know that)

density of water = 1g/ml

density = mass/volume

1g/ml = M(w)/1982ml

M(w) = 1982g

C(w) = specific heat capacity of water = 4.184J/g-°C

∆T = change in temperature of water = 21.4 -25 = -3.6°C

putting the all value in formula

(X)(0.502J/g°C)(25.4°C) + (1982g)(4.148J/g°C)(-3.6°C) = 0

(12.7508J/g)(X) + (-29853.6768J ) = 0

X = (29853.6768J)/(12.7508J/g) = 2341.318g

So mass of stainless steel = 3241.318g

(If you have other specific heat and density of water then you put these values and solve like this)

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