How do I calculate specific heat capacity in a lab report?

[soccerLuver]

Homework Statement

Homework Equations

i have a problem solving this foe my lab report..i just don't get the formula
heres the data:
ethylene glycol...its mass is 200 g ,initial temperature 22C, temperature after heated 100C
what is its heat capcity?

metal and water
mass of metal-15.8 g
temperature of water-23 C
temperature of metal and washers-24 C
what is the heat gained by the water assuming the heat gained by the water is equal to the heat lost by the metal?
what is the heat capacity of the metal?

any clue..or formula?

The Attempt at a Solution

200( 1cal/gC) (1) / 15.8 (1)

 

[Borek]

In general, formula that you are looking for is most likely

q=c*m*ΔT

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methods

 

[Blueshift5]

(1) = 12.66 cal/gC

To solve these specific heat problems, we need to use the formula Q = mcΔT, where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. For the first problem with ethylene glycol, we can calculate the heat capacity by plugging in the given values: Q = (200 g)(1 cal/gC)(100C - 22C) = 15,600 cal. Then, we can use the formula c = Q/mΔT to find the specific heat capacity: c = (15,600 cal)/(200 g)(78C) = 0.995 cal/gC.

For the second problem with the metal and water, we can use the fact that the heat gained by the water is equal to the heat lost by the metal. This means that Qwater = Qmetal, so we can set up the equation (mwater)(cwater)(ΔTwater) = (mmetal)(cmetal)(ΔTmetal). We are given the mass and temperature changes for both substances, so we can solve for the heat gained by the water: Qwater = (15.8 g)(1 cal/gC)(1C) = 15.8 cal. This is also equal to the heat lost by the metal, so we can use the same formula as before to find the heat capacity of the metal: (15.8 cal)/(15.8 g)(1C) = 1 cal/gC.

In summary, to solve specific heat problems, we need to use the formula Q = mcΔT and rearrange it as needed to solve for the specific heat capacity. It is also important to remember that the heat gained by one substance is equal to the heat lost by another substance, as seen in the second problem. I hope this helps with your lab report!