A 33.1 g ring that appears to be silver is heated to a temperature of 77°C, and then placed in a calorimeter containing 5.00 10-2 kg of water at 24.0°C. The calorimeter is not perfectly insulated, however, so that 0.14 kJ of energy is transferred by heat to the surroundings by the time an equilibrium temperature of 25.0°C is reached.
Determine the specific heat capacity of the ring. Watch the units!
To determine the specific heat capacity of the ring, we can use the equation:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred between the ring and the water using the equation q = mcΔT.
The mass of the water, m_water, is 5.00 × 10^-2 kg.
The change in temperature, ΔT, is (25.0°C - 24.0°C) = 1.0°C = 1.0 K.
Next, let's calculate the heat transferred to the surroundings.
The heat transferred to the surroundings, q_surroundings, is given in the problem as 0.14 kJ. We need to convert it to joules since the specific heat capacity has units of J/g°C.
1 kJ = 1000 J, so 0.14 kJ = (0.14 kJ) * (1000 J/kJ) = 140 J.
Now, let's rearrange the equation q = mcΔT to solve for the specific heat capacity, c:
c = q / (m_ring * ΔT)
We have the values of q_surroundings and ΔT, and we have the mass of the ring, m_ring, which is given as 33.1 g. We need to convert the mass from grams to kilograms to match the units of the mass of water.
1 g = 0.001 kg, so 33.1 g = (33.1 g) * (0.001 kg/g) = 0.0331 kg.
Now we can calculate the specific heat capacity:
c = 140 J / (0.0331 kg * 1.0 K)
c = 4215.59 J/(kg*K)
Therefore, the specific heat capacity of the ring is approximately 4215.59 J/(kg*K).