A 150.0-g sample of a metal at 27.9°C is added to 150.0 g of H2O at 13.1°C. The temperature of the water rises to 15.7°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.
Solve this equation for the specific heat capacity of the metal (H_metal), then plug in the values
Ok so, 150 + x + (15.7-27.9) + ( 150 * ?* (15.7- 13.1) ) = 0What would the H of H20 be ?
The specific heat of water is (4.18 J/g*degrees C)
Ok thanks , oh and the beginning part of the formula is multiplication not addition
Ah yeah it is apologies
Ok for the ice problem it's the same calculation, but I simplified this one, just make sure to carry the negative signs properly to get the right answer.
Thanks sooo much , I really appreciate it.
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?