Discover the Heat Transfer in a Thermo-Aluminum Calorimeter with Metallic Blocks

[PrudensOptimus]

An aluminum calorimeter of mass 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 12.0°C. Two metallic blocks are placed into the water. One is a 40.0 g piece of copper at 80.0°C; the other block has a mass of 13.0 g and is originally at a temperature of 116.0°C. The entire system stabilizes at a final temperature of 15.0°C.



Here's what I have so far:

(mcΔT)_A + (mcΔT)_W = (mcΔT)_Cu + (mcΔT)_Unknown

 

[Evalesco]

100(3) + 250(3) = 40(67) + 13(103)

300 + 750 = 2710 + 1339

1050 = 4049

It looks like the total heat lost by the two blocks is 4049 J. Is that correct? What is the mass and final temperature of the unknown block?

 

[M98Ranger]

Where:
m = mass
c = specific heat capacity
ΔT = change in temperature
A = aluminum calorimeter
W = water
Cu = copper block
Unknown = unknown metallic block

To solve for the specific heat capacity of the unknown metallic block, we can rearrange the equation to:

cUnknown = [(mcΔT)A + (mcΔT)W - (mcΔT)Cu] / (mΔT)Unknown

Plugging in the given values, we get:

cUnknown = [(100g)(0°C - 15°C) + (250g)(0°C - 15°C) - (40.0g)(80.0°C - 15.0°C)] / (13.0g)(116.0°C - 15.0°C)

cUnknown = -15000 J / (13.0g)(101.0°C)

cUnknown = -11.65 J/g°C

This negative value suggests that the unknown metallic block has a lower specific heat capacity than the other materials in the system, meaning it can absorb heat more easily. This could indicate that the unknown block is made of a metal with a lower atomic mass, such as silver or gold. Further experiments and calculations would be needed to confirm this hypothesis.