Question

I NEED HELP ASAP PLS

A 638.3 g sample of nickel releases 12337 joules of heat. The specific heat capacity of nickel is
0.5024 J/(g.°C). By how much did the temperature of this sample change, in degrees Celsius?

A 0.6949 mol sample of aluminium absorbs 5281.1 joules of heat. The specific heat capacity of aluminium is 24.2 J/(mol K). By how much did the temperature of this sample change, in kelvins?

A 0.4363 mol sample of indium experiences a temperature change of +109.71 °C while absorbing 1279.9 joules of heat. What is the specific heat capacity of indium?

A 824.5 g sample of silver undergoes a temperature change of +1078.87 K while absorbing
207260 joules of heat. What is the specific heat capacity of silver?

A 616.7 g sample of iridium is subjected to a temperature change of -2020.47 °C while releasing 1617300 joules of heat. What is the specific heat capacity of iridium?

A 875.8 g sample of uranium absorbs 15291 joules of heat. The specific heat capacity of uranium is 0.116 J/(g K). By how much did the temperature of this sample change, in kelvins?

A 990.1 g sample of molybdenum absorbs 179910 joules of heat. The specific heat capacity of molybdenum is 0.2772 J/(g.°C). By how much did the temperature of this sample change, in degrees Celsius?

A 3.596 mol sample of barium is subjected to a temperature change of -568.85 K. The specific heat capacity of barium is 28.07 J/(mol K). How many joules of heat were transferred by the sample?

A sample of rhodium experiences a temperature change of -1311.77 °C while releasing 203580 joules of heat. The specific heat capacity of rhodium is 0.2428 J/(g.°C). What is the mass of this sample?

A sample of methane goes through a temperature change of +330.22 °C while absorbing 672220 joules of heat. The specific heat capacity of methane is 2.191 J/(g.°C). What is the mass of this sample?

Answer 1

For the first problem, using the formula Q = mcΔT, we can solve for ΔT, where Q is the heat released, m is the mass of nickel, c is the specific heat capacity of nickel, and ΔT is the temperature change. Plugging in the values, we get ΔT = Q/(mc) = 12337/(638.3 × 0.5024) ≈ 48.9 °C.

For the second problem, we use the same formula, but this time we solve for ΔT in kelvins, since the specific heat capacity of aluminum is given in units of J/(mol K). Plugging in the values, we get ΔT = Q/(mc) = 5281.1/(0.6949 × 24.2) ≈ 97.8 K.

For the third problem, we use the formula Q = mcΔT, but this time we solve for c, since the specific heat capacity of indium is unknown. Plugging in the values, we get c = Q/(mΔT) = 1279.9/(0.4363 × 109.71) ≈ 27.8 J/(mol K).

For the fourth problem, we use the same formula, but this time we solve for the specific heat capacity of silver. Plugging in the values, we get c = Q/(mΔT) = 207260/(824.5 × 1078.87) ≈ 0.240 J/(g K).

For the fifth problem, we use the same formula to solve for the specific heat capacity of iridium. Plugging in the values, we get c = Q/(mΔT) = -1617300/(616.7 × -2020.47) ≈ 0.131 J/(g K).

For the sixth problem, we use the same formula to solve for the temperature change in kelvins. Plugging in the values, we get ΔT = Q/(mc) = 15291/(875.8 × 0.116) ≈ 149.8 K.

For the seventh problem, we use the same formula to solve for the temperature change in degrees Celsius. Plugging in the values, we get ΔT = Q/(mc) = 179910/(990.1 × 0.2772) ≈ 67.5 °C.

For the eighth problem, we use the formula Q = mcΔT to solve for the heat transferred. Plugging in the values, we get Q = mcΔT = 3.596 × 28.07 × -568.85 ≈ -5789 J.

For the ninth problem, we use the same formula to solve for the mass of rhodium. Plugging in the values, we get m = Q/(cΔT) = 203580/(0.2428 × -1311.77) ≈ 197.8 g.

For the tenth problem, we use the same formula to solve for the mass of methane. Plugging in the values, we get m = Q/(cΔT) = 672220/(2.191 × 330.22) ≈ 112.4 g.