A 100-g piece of metal initially at T = 75oC

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SUMMARY

The problem involves a 100-g piece of iron at an initial temperature of 75°C submerged in 100 g of water at 25°C. Using the specific heat capacities of iron (0.45 J g-1 °C-1) and water (4.18 J g-1 °C-1), the final temperature can be calculated using the equation q=mc(delta T). The correct approach requires setting the heat lost by the metal equal to the heat gained by the water, leading to the equation 100g * 0.45 J g-1 °C-1 * (75°C - Tfinal) = 100g * 4.18 J g-1 °C-1 * (Tfinal - 25°C).

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Homework Statement


A 100-g piece of metal initially at T = 75oC is submerged in 100 g of water initially at T = 25oC. The specific heat capacity of iron is 0.45 J g-1 °C-1 and the specific heat capacity of water is 4.18 J g-1 °C-1. What is the final temperature of both substances in oC?


Homework Equations


q=mc(delta T)


The Attempt at a Solution


well i tried to set mc(delta T)=mc(delta T), plug in all the values and solve for T(final), but i keep getting the wrong answer
 
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Can you show your work? It is hard to see what your problem is without being able to check your calculations.
 

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