The mass of the metal is 45.00 g. The metal's highest temperature is 82 degrees C.
With the information above and the specific heat of water, 4.184 J/(g·°C), calculate the specific heat, c, of the metal.
idk if this is useful or not, but the mass of water is 130.00 g and the initial temp of water was 26 degrees C and the final temp was 30 degrees C
With the information above and the specific heat of water, 4.184 J/(g·°C), calculate the specific heat, c, of the metal.
idk if this is useful or not, but the mass of water is 130.00 g and the initial temp of water was 26 degrees C and the final temp was 30 degrees C
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Here's my understanding: You dropped a 45-g piece of metal at 82 C into 130 g of H2O at 26 C and the final temperature was 30 C.
Heat gained by H2O = (mass H2O)(SH H2O)(Tf - Ti) = (130.00 g)(4.184 J/g C)(30 C - 26 C) = 2180 J
Heat lost by metal = 2180 J = (mass metal)(SH metal)(Ti - Tf)
2180 = (45.00 g)(SH metal)(82 C - 30 C)
2180 = 3690 SH metal - 1350
3530 = 3690 SH metal
SH metal = 3530 / 3690 = 0.96 J/g C
Heat gained by H2O = (mass H2O)(SH H2O)(Tf - Ti) = (130.00 g)(4.184 J/g C)(30 C - 26 C) = 2180 J
Heat lost by metal = 2180 J = (mass metal)(SH metal)(Ti - Tf)
2180 = (45.00 g)(SH metal)(82 C - 30 C)
2180 = 3690 SH metal - 1350
3530 = 3690 SH metal
SH metal = 3530 / 3690 = 0.96 J/g C