When 47.5 J of heat are added to 13.2 g of a liquid, its temperature raises by 1.72 degree Celsius. What is the heat capacity of the liquid?
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Originally Posted by Louisana1 When 47.5 J of heat are added to 13.2 g of a liquid, its temperature raises by 1.72 degree Celsius. What is the heat capacity of the liquid? I think that if you were to look up the definition of "heat capacity" that would make what you need to do clear.
I am still having issues placing these numbers in the correct spot. Is there anyone that can point me in the right direction, or share some links that would help me with this further.
$\displaystyle Q = m \cdot C \cdot \Delta T$ you are given $\displaystyle Q$ , $\displaystyle m$ , and $\displaystyle \Delta T$ ... solve for $\displaystyle C$
47.5J(13.2g)(4.18j)(1.72c)= 341.506 grams signif digits 341.5 Is this right?
Q=m·c·Δt 47.5 joules = 13.2g·4.18Joules·1.72°c 47.5 = 94.90272g Do I divide this I'm not sure if this is right?Can someone point me in the right direction?
$\displaystyle Q = m \cdot C \cdot \Delta T$ $\displaystyle \frac{Q}{m \cdot \Delta T} = C$ substitute in your given values and calculate the specific heat, C.
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