Q1) Copper has a specific heat of 385 Jkg-1K-1. How much heat is lost when 25.0

Question

Q1) Copper has a specific heat of 385 Jkg-1K-1. How much heat is lost when
25.0 g of copper cools from a temperature of 415.0oC to a temperature of
22.0 oC?

a)The temperature of a water sample increases by 50 oC when heated by
22990 J. The specific heat of liquid water is 4180 Jkg-1K-1. What is the mass of the sample of water? Assume no heat loses to the surroundings.

b)A piece of unknown metal with mass 50 g is heated to 160.65 oC and
dropped into 110 g of water at 20 °C. The final temperature of the system is 32.5 °C. Calculate the specific heat of the metal. The specific heat of liquid water is 4180 Jkg-1K-1. Assume no heat loses to the surroundings.

Explanation / Answer

Q1) Copper specific heat : S = 385 Jkg-1K-1 Mass of the copper = m = 25 x10-3kg Temparature difference : T = (415.0-22.0) = 393 oC heat is lost : Q = m S T = ( 25 x10-3)(385 )(393)                                            = 3782.625 J (a) specific heat of liquid water is: S = 4180 Jkg-1K-1 temperature of a water sample increases by= T = 50 oC Heat given to the water : Q = 22990 J Mass o the water sample = m = ? Q = mS T m = Q / S T       = (22990)/( 4180 )(50)       = 0.11 Kg (b) Mass of the Metal = m = 50g = 50 x10-3kg Metal temparature : t = 160.65oC Mass of the water = m' = 110 g = 110x10-3Kg water temparature : t ' = 20oC   final temperature of the system is : T =32.5 °C   specific heat of liquid water is = S ' = 4180 Jkg-1K-1 specific heat of the metal = S = ? Acc. to thermal equilibruim , Heat lost by the metal = Heat gained by the water Qm = Qw m S (160.65 - 32.5 ) = m' S' (32.5 - 20) (50 x10-3)(S)(128.15) = (110x10-3)( 4180 )(12.5)                    6.4075 ( S  ) = 5747.5 thus,     specific heat of the metal    = S = 896.99                        or                         S ˜ 897  Jkg-1K-1                        or                         S ˜ 897  Jkg-1K-1     

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---