1. calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × δt.

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.

the water has absorbed the heat of the metal. so, qwater = qmetal
using the formula qmetal = m × c × δt, calculate the specific heat of the metal. use the data from your experiment for the metal in your calculation.

(for the first image's data)

2. calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × δt.

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. show all your work.

the water has absorbed the heat of the metal. so, qwater = qunknown metal. using the formula qunknown metal = m × c × δt, calculate the specific heat of the metal. use the data from your experiment for the unknown metal in your calculation. show all your work.

(for the second image's data)

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

Q2:  Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.  

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. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.

a) First part: the energy change (q) of the surroundings (water):

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (25 g)(4.18 J/g°C)(6.4°C) = 668.8 J.

b) second part:

Q water = Q unknown metal.

Q unknown metal =  - 668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

Q unknown metal =  - 668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

- 668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (- 668.8)/(- 1914) = 0.3495 J/g°C.

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

: )

the temperature of the gas, in k, not celsius.