If 0.241mol of ammonium chloride NH4Cl is dissolved in a liquid producing ×−2.43102cal of heat, calculate the temperature change of ×6.51102g of the resulting solution if the specific heat capacity of the solution is 0.800cal·g°C. Round your answer to 3 significant figures.
Question
If 0.241mol of ammonium chloride NH4Cl is dissolved in a liquid producing ×−2.43102cal of heat, calculate the temperature change of ×6.51102g of the resulting solution if the specific heat capacity of the solution is 0.800cal·g°C. Round your answer to 3 significant figures.
Solution
To solve this problem, we need to use the formula for heat transfer:
q = mcΔT
where: q = heat absorbed or released m = mass c = specific heat capacity ΔT = change in temperature
We know that the heat released is -2.4310^2 cal, the mass of the solution is 6.5110^2 g, and the specific heat capacity is 0.800 cal/g°C. We want to find ΔT, the change in temperature.
Rearranging the formula to solve for ΔT gives us:
ΔT = q / (mc)
Substituting the given values into the formula gives us:
ΔT = -2.4310^2 cal / (6.5110^2 g * 0.800 cal/g°C)
Performing the calculation gives us:
ΔT = -0.595°C
So, the temperature change of the solution is -0.595°C. The negative sign indicates that the temperature of the solution decreased.
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