Question

A silver block, initially at 55.4 ∘C, is submerged into 100.0 g of water at 24.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.8 ∘C. What is the mass of the silver block? Express your answer to two significant figures and include the appropriate units.

Answer 1

Hi Dear Friend this is essentially Heat balance problem. Please note that you have not provided specific heat capacities of silver and water, so i am using following values, if in case they do not match with your values please report that in comments section below, so that i will re-calculate it again for you.

water 1 cal = 4.184 J/g^oC; silver 0.058 cal = 0.242672 J/g^oC

heat gained by 100 g A (water) = heat lost by y g B (silver)

So m_A x C_A x temp. Diff. Of A = m_B x C_B x temp diff of B

m_A x C_A x (T_final - T_initial) = m_B x C_B x (T_initial - T_final)

Let m_B be the mass of B

4.184 J/g^oC x 100 g x ( 27.8 - 24) ^oC = 0.242672 J/g^oC x y g x ( 55.4 - 27.8) ^oC

y = [4.184 J/g^oC x 100 g x ( 27.8 - 24) ^oC] / [0.242672 J/g^oC x ( 55.4 - 27.8) ^oC]

y = [1589.92 J] / [6.6977472 J/ g]

y = 237.381309345 g = 240 grams

mass of the silver block = 240 g

Answer : mass of the silver block = 240 g (2 sig fig)

Hope this helped you!

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