Answer:
Problem 1 => 8778 joules
Problem 2 => 14,630 joules
Step-by-step explanation:
Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.
For the listed problems...
Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.
Note the temperature change in the problem context => use q = m·c·ΔT ...
q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)
Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.
Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...
q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules
Hope this helps. Doc