Question

A 71.5 g piece of metal at 81.0°C is placed in 129 g of water at 21.0°C contained in a calorimeter. The metal and water come to the same temperature at 24.3°C. How much heat (in J) did the metal give up to the water? (Assume the specific heat of water is 4.18 J/g·°C across the temperature range.)

__________J
What is the specific heat (in J/g·°C) of the metal?
__________J/g·°C
check your answer for the correct number of significant figures.

Answer 1

The metal gave up 1780 J of heat to the water. Its specific heat capacity, calculated using the mass and temperature change of the metal and the heat absorbed by the water, is 0.437 J/g°C.

Step-by-step explanation:

To find the heat that the metal gave up to the water, we can use the formula 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.

The water's temperature increased from 21.0°C to 24.3°C, so ΔT for the water is 3.3°C. The mass of water is 129 g, and the specific heat capacity is 4.18 J/g°C.

First, calculate the heat absorbed by the water: q = (129 g)(4.18 J/g°C)(3.3°C) = 1775.634 J. This heat came from the metal, so the metal lost the same amount of heat.

To find the specific heat capacity of the metal, we'll assume that the heat lost by the metal (q) is equal to the heat gained by the water. The metal's mass is 71.5 g and ΔT for the metal is (81.0°C - 24.3°C) = 56.7°C.

Since q for metal must equal -q for water (heat lost is negative): q = (71.5 g)(c)(-56.7°C) = -1775.634 J. We can now solve for specific heat capacity c.

‑c = 1775.634 J / (71.5 g ‑ 56.7°C) c = 0.437 J/g°C

We round our answers to the least number of significant figures used in the measurements. For the heat given up by the metal, the least number of significant figures is 3, so we round to 1780 J. For the specific heat capacity of the metal, the less accurate measurement is the mass of the metal with 3 significant figures, so we round the specific heat capacity to 0.437 J/g°C.