Question

THERMAL EQUILİBRİUMAȚWORKOUT, PART 1 Consider a mixture of two substances in an insulated styrofoam container such as one in the photo below and the diagram below thermometer thermorneter Substance H Substance R A drop of substance R in a styrofoam container full of substance H, with a styro- foam lid back on tight and a thermometer in place. The drop of substance R has a mass of 102 grams. Its initial temperature is TR -1000 K The mass of substance H is 83 grams, and it has initial temperature TH -1100 K Here are the thermal specs for these two substances Substance R A drop of substance R in a styrofoam container full of substance H, with a styro- foam lid back on tight and a thermometer in place. The drop of substance R has a mass of 102 grams. Its initial temperature is TR-1000 K. The mass of substance H is 83 grams, and it has initial temperature T, 1100 K. Here are the thermal specs for these two substances Substance Specific heat (cal/g K) Melting point (K) 0.336 807 0.828 900 QUESTION: When substance R absorbs 100 calories from substance H, substance R heats up by how many K? Calculate ΔΤ R to the nearest 0.1K. E.g., if your answer is Δ1R-5558 K, then type 5 . 6 in the answer box.

Answer 1

If the substance R absorbs 100 calories of heat:

Heat absorbed is given by: m*s*(Tf - Ti)

m=102grams, s=0.336cal/g K, Ti=1000K.

Thus, 100= 102*0.336*(Tf - 1000)

Thus, Tf - 1000= 100/34.27 = 2.92

Thus, Tf = 1002.92 K = 1002.9 K