Ablock of copper of unknown mass has an initial temperature of 65.4 ∘c. the copper
Question:
Answers
37.840g
Explanation:
We have to use this equation:
Q=m*c*ΔT
Q= Symbol of heat transfer
ΔT= Change in temperature
c = specific heat.
First, we calculate the amount of heat transferred to water from copper:
m = 95.7g
c = 4.1813 J/gºC (It's a known value)
ΔT = 24.2ºC - 22.7ºC = 1.5
Q = (95.7g)(4.1813J/gºC)(1.5ºC)
Q = 600.226J This is the heat absorbed by water which is the same released by copper
So Q water = - Q copper
Now we can calculate the mass of copper like this:
Q = - 600.226J
c = 0.385J/gºC
ΔT = (24.2ºC - 65.4ºC) = -41.2ºC
m = ?
we solve for m in the equation Q=m*c*ΔT and we find:
[tex]\frac{Q}{c ΔT} = m[/tex]
[tex]\frac{-600.226J}{(0.385J/gºC) (- 41ºC)} = 37.840g[/tex]
The mass of the copper block: 37,829 g
Further explanationThe law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Q in = Q outHeat can be calculated using the formula:
[tex]\rm \large{\boxed{\bold{Q=m\times c\times \Delta T}}}[/tex]
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
1. H₂Om = 95.7 g
c = 4.18 J / gºC
delta T = 24.2 °C - 22.7 °C = 1.5 °C
Q H₂O = 95.7 x 4.18 x 1.5
Q H₂O = 600,039 J
2. Cuc = 0.385 J / gºC
delta T = 65.4 °C - 24.2 °C = 41.2 °C
QH₂O = Q Cu
600,039 J = m. 0.385 J / gºC. 41.2. °C
[tex]\rm m=\dfrac{600.039}{0.385\times 41.2}\\\\m=\boxed{37.829\:g}}[/tex]
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