Question

1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

(27.776 g) * ( 4.18 J/g *C ) * ( 25.3 C )
Q=2944kJ
^^ is this correct?
2. Calculate the specific heat of the metal.

Answer 1

use the formula: q=m*C*dT where dT=T(final)-T(initial)

Answer 2

(27.776 g) * ( 4.18 J/g *C ) * ( 25.3 C )
Q=2944kJ

Is this correct

Answer 3

If you plug it in your calculator and that is your answer, then you should be correct.

Answer 4

you were given the change in temperature, dT, as 25.3?

Answer 5

Honestly I just need someone to walk me through this

Answer 6

Measured Mass of Metal: 27.776g
Distilled Water Volume: 26 mL
Distilled Water Temperature: 25.3 *C
Temperature of Metal: 100.5*C
Temperature of Mixture:38.9*C


1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
qwater = m × c × ΔT

m = mass of water = density x volume = 1 x 26 = 26 grams

ΔT = T(mix) - T(water) = 38.9 - 25.3 = 13.6 

q(water) = 26 x 13.6 x 4.18 

q(water) = 1478 Joules


2. Calculate the specific heat of the metal.
3. q(water) = - q(metal)
q(metal) = - 1478 Joules
q(metal) = m × c × ΔT
m = 27.776 g
ΔT = T(mix) - T(metal) 
ΔT = 38.9 - 100.5 = - 61.6 
C = q(metal) / m x ΔT
C = -1478 / (-61.6 x 27.776 )
C = 0.864 J / (g × °C)

Answer 7

idk if I'm doing this correctly

Answer 8

its good (Y)