When 31 g of a metal at 88 ◦C is added to
46 g of water at 27 ◦C, the temperature of the water rises to 34 ◦C. What is the specific heat capacity of the metal? Assume no heat was lost to the surroundings. Answer in units of J / g ·◦C.
To calculate the specific heat capacity of the metal, we can use the formula:
q = m × c × ΔT
where:
q is the heat absorbed or released by the substance,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature of the substance.
In this case, the metal is absorbing heat from the water, so the heat gained by the metal (qmetal) is equal to the heat lost by the water (qwater). We can set up the equation as follows:
qmetal = qwater
Since we assume no heat was lost to the surroundings, we can calculate qwater using the formula:
qwater = mwater × cwater × ΔTwater
where:
mwater is the mass of water,
cwater is the specific heat capacity of water,
ΔTwater is the change in temperature of water.
We are given the following values:
mwater = 46 g
cwater = 4.18 J/g·°C (specific heat capacity of water)
ΔTwater = 34°C - 27°C = 7°C
Plugging in these values, we can calculate qwater:
qwater = (46 g) × (4.18 J/g·°C) × (7°C) = 1349.08 J
Since qmetal is equal to qwater, we can now calculate the specific heat capacity of the metal (cmetal). We rearrange the formula:
qmetal = mmetal × cmetal × ΔTmetal
Dividing both sides by mmetal × ΔTmetal:
cmetal = qmetal / (mmetal × ΔTmetal)
We are given the following values:
mmetal = 31 g
ΔTmetal = 34°C - 88°C = -54°C
Plugging in these values, we can calculate cmetal:
cmetal = 1349.08 J / (31 g × -54°C) ≈ -0.828 J/g·°C
Therefore, the specific heat capacity of the metal is approximately -0.828 J/g·°C.