A50.0 gram sample of a metal initially at a temperature of 200 degrees celsius is immersed in 400 gram of water at 20 degrees celsius. find the specific heat of the metal in j/kg degrees celsius, if the equilibrium temperature is 23 degrees celsius. (specific heat of water is 4186 j/kg degrees celsius)

Specific heat of metal,

Given:

Mass of metal, M = 50.0 g = 0.05 kg

Initial temperature, T =

Mass of water, M' = 400 g = 0.4 kg

Temperature for water, T' =

Equilibrium temperature,

specific heat of water,

Solution:

At equilibrium:

specific heat of metal = specific heat of water

Therefore,

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

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