If the specific heat of water = 4190 J/kg /°C and a total of 120g of hot water at 95°C is poured into the calorimeter, resulting in a final equilibrium temperature of 73°C, how many joules of heat are lost by the water? 11061.6


If the specific heat of copper = 390 J/kg /°C and the mass of the copper calorimeter is 70 g , how many joules of heat are gained by the calorimeter as its temperature changes from 22°C to the equilibrium temperature of 73°C? 11392.3

Need help with this part -
If the water had lost 12000 joules and the calorimeter gained 1300 joules as the temperature of both rock and calorimeter increased from 22°C to 73°C and the mass of rock in the calorimeter was 185 grams, what is the specific heat, c, of the rock material?
(CALCULATE YOUR ANSWER IN SI UNITS, but do not state the units in the answer).
Answer

Heat lost by water equals

(Heat gained by calorimeter)
PLUS
(Heat gained by rock)

12,000 = 1300 + Mrock*Crock*(delta T)

10,700= 0.185*Crock*51

Solve for Crock in J/kg*C

thank you

To find the specific heat, c, of the rock material, we can use the formula:

Q = mcΔT

Where:
Q is the heat energy gained or lost (in joules)
m is the mass of the substance (in kg)
c is the specific heat capacity of the substance (in J/kg/°C)
ΔT is the change in temperature (in °C)

In this case, the water lost 12000 joules of heat and the calorimeter gained 1300 joules. So, the total heat energy gained or lost is:

Q_total = -12000 J + 1300 J = -10700 J

The mass of the rock in the calorimeter is 185 grams, which is equivalent to 0.185 kg.

The temperature change, ΔT, is equal to the final temperature minus the initial temperature:

ΔT = 73°C - 22°C = 51°C

Now we can rearrange the formula to solve for c:

c = Q_total / (m * ΔT)

c = -10700 J / (0.185 kg * 51°C)

c ≈ -10700 J / (0.185 kg * 51 K)

c ≈ -10700 J / 9.435 kg K

c ≈ -1134 J/kg/K

So, the specific heat, c, of the rock material is approximately -1134 J/kg/K.