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Sup_open~study
 3 years ago
can some one show me how to work this out and what to do please thank you.
The specific heat ( C) of water is quite high compared to the specific heat of lead, so the energy released by a sample of heated lead to its surroundings will only around 4% of the amount of heat released by an equal mass of water under identical conditions.
(a) Calculate the change in temperature that results from the addition of 2500 j of heat energy to a 25g sample of water .
Sup_open~study
 3 years ago
can some one show me how to work this out and what to do please thank you. The specific heat ( C) of water is quite high compared to the specific heat of lead, so the energy released by a sample of heated lead to its surroundings will only around 4% of the amount of heat released by an equal mass of water under identical conditions. (a) Calculate the change in temperature that results from the addition of 2500 j of heat energy to a 25g sample of water .

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Sup_open~study
 3 years ago
Best ResponseYou've already chosen the best response.0@Sheng can you help me plz

JFraser
 3 years ago
Best ResponseYou've already chosen the best response.1You need to have one of two things: the specific heat capacity of water, or of lead. It's easy to look up both, but water is the more useful one, since it's water that you're heating up. The equation used for heat transfer is:\[Q = m*C_P*\Delta T\] Q is the heat released (or absorbed), m is the mass of the stuff doing the absorbing (in this case, water), C_P is the specific heat of that substance (4.184J/gC for water), and DT is the change in temp, which is what you're looking for Plug and solve.
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