# In a calorimeter, a 90.0 g piece of metal at 92.4 *C is submerged in 100.0 g of water, that is initially at 21.6 *C. When the metal and water reach thermal equilibrium, the temperature is 29.3 *C. What was the heat capacity of the metal (3 sig figs)?

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In a calorimeter, a 90.0 g piece of metal at 92.4 *C is submerged in 100.0 g of water, that is initially at 21.6 *C. When the metal and water reach thermal equilibrium, the temperature is 29.3 *C. What was the heat capacity of the metal (3 sig figs)?

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Step 1

In a calorimeter, a 90.0 g piece of metal at 92.4 °C is submerged in 100.0 g of water that is initially at 21.6 °C. When the metal and water reach thermal equilibrium, the temperature is 29.3 °C. The heat capacity of the metal (3 sig figs) is to be calculated.

Step 2

In a calorimeter, heat gained by one substance is equal to the heat lost by the other substance.

Where

Heat (q) gained or lost is expressed as:

q = m×c×ΔT

Where

m- Mass of the substance

c- Specific heat of the substance

ΔT – Change in temperature

Step 3

Therefore,

Heat lost by the metal = heat gained by the water

mm×cm×ΔTm= -...

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